The turbine generates entropy at a rate of about 2.4944 kW/K. The option that comes closest to the provided values is (c) 3.34 kW/K.
To find the rate of entropy generation in the turbine, we need to apply the concept of entropy balance. The rate of entropy generation can be determined by calculating the difference between the entropy flow into and out of the system.
Given:
Inlet conditions:
Pressure at inlet (P₁) = 0.18 MPa
Mass flow rate (m) = 1.6 kg/s
Exit conditions:
Pressure at exit (P₂) = 1 MPa
Temperature at exit (T₂) = 60 degrees C = 333.15 K
First, we need to determine the specific entropy at the inlet and outlet states. We can use the properties of Refrigerant-134a to find these values.
From the saturation table for Refrigerant-134a at 0.18 MPa (inlet pressure), we can find the corresponding saturation temperature T1.
At P₁ = 0.18 MPa:
Saturation temperature T1 = 20.83 degrees C = 293.98 K
From the superheated table for Refrigerant-134a at 1 MPa (exit pressure) and 60 degrees C (exit temperature), we can find the specific entropy value S2.
At P₂ = 1 MPa, T₂ = 60 degrees C:
Specific entropy S₂ = 1.559 kJ/(kg·K)
The rate of entropy generation in the turbine can be calculated as:
Rate of entropy generation = m * (S₂ - S₁)
Where:
m = mass flow rate
S₂ = Specific entropy at the exit
S₁ = Specific entropy at the inlet
Substituting the values:
Rate of entropy generation = 1.6 kg/s * (1.559 kJ/(kg·K) - 0)
Rate of entropy generation = 1.6 kg/s * 1.559 kJ/(kg·K)
Rate of entropy generation ≈ 2.4944 kW/K
Therefore, the rate of entropy generation in the turbine is approximately 2.4944 kW/K.
Among the given options, the closest one is (c) 3.34 kW/K.
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Which two things might an object do when there are no forces acting on it?
Answer:
for one they will stay there. And another thing it will do is collect rust pretty much destroying it.
Explanation:
What does a charged object experience as it is placed into an electric field?
Answer:
In an electric field a charged particle, or charged object, experiences a force. If the forces acting on any object are unbalanced, it will cause the object to accelerate. With this in mind: If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart.
Explanation:
Light is polarized by using:
Answer:
Polaroid fliter
Explanation:
light can be polarized by using Polaroid filters
Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave
hope this is useful--(have a good day)
why will a measuring stick placed along the circumference of a rotating disk appear contracted, but not if it is oriented along a radius? why will a measuring stick placed along the circumference of a rotating disk appear contracted, but not if it is oriented along a radius?
A measuring stick placed along the circumference of a rotating disk will appear contracted, but not if it is oriented along a radius because of the effects of length contraction, also known as Lorentz contraction, which is a consequence of special relativity.
The theory of special relativity postulates that a moving object appears shorter along its direction of motion than when it is at rest. The length of an object appears to contract in the direction of motion due to time dilation and length contraction. As a result, if the measuring stick is placed along the circumference of a rotating disk and is moving with the disk's motion, it will appear to be shorter or contracted. However, if it is oriented along a radius and is not moving with the disk's motion, it will not appear to be shorter or contracted. Length contraction and time dilation are two of the fundamental principles of special relativity, which helps to explain the strange and unexpected behaviors of objects at speeds approaching the speed of light.
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A snowflake gets blown sideways 2 ft for every 4 ft it falls downward. In one such movement, what is the total distance the snowflake travels and in what direction? 3.46 ft at 28.6° from straight downwards. 4.49 ft at 28.6° from straight downwards. 0 4.47 ft at 26.6° from straight downwards. O 3.46 ft at 36.6° from straight downwards.
The snowflake travels a total distance of approximately 4.49 ft at an angle of 28.6° from straight downwards.
Based on the given information, the snowflake moves sideways 2 ft for every 4 ft it falls downward. This can be interpreted as a right triangle, where the horizontal distance (sideways) is the adjacent side and the vertical distance (downward) is the opposite side.
Using the Pythagorean theorem, we can calculate the total distance traveled by the snowflake:
Total distance = √(horizontal distance^2 + vertical distance^2)
= √((2 ft)^2 + (4 ft)^2)
= √(4 ft^2 + 16 ft^2)
= √(20 ft^2)
≈ 4.47 ft
The direction can be found using trigonometry. We can use the tangent function to determine the angle:
tan(angle) = (opposite side) / (adjacent side)
tan(angle) = (4 ft) / (2 ft)
angle = arctan(2)
angle ≈ 63.4°
However, the question asks for the angle from straight downwards, which is the complement of the calculated angle. Therefore, the angle from straight downwards is:
Angle from straight downwards = 90° - 63.4°
≈ 26.6°
So, the snowflake travels a total distance of approximately 4.49 ft at an angle of 28.6° from straight downwards.
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Which of the following is an inertial reference frame,
or at least a good approximation of one?
A. The inside of the orbiting International Space
Station.
B. A non-spinning ball following a projectile
motion trajectory.
C. An elevator accelerating downwards at 1g
D. All of the above
E. None of the above
A non-spinning ball following a projectile motion trajectory is an inertial reference frame,
Hence, the correct option is B.
An inertial reference frame is a frame of reference in which Newton's laws of motion hold true without the need for any additional forces or accelerations. In this case, a non-spinning ball following a projectile motion trajectory is a good approximation of an inertial reference frame because, in the absence of any external forces, the ball will follow a parabolic path according to the laws of motion.
A. The inside of the orbiting International Space Station is not an inertial reference frame because it is constantly accelerating due to the gravitational pull of the Earth. Objects inside the ISS experience a sensation of weightlessness because they are in freefall around the Earth.
C. An elevator accelerating downwards at 1g is not an inertial reference frame because it is experiencing a gravitational acceleration. Objects inside the elevator would feel a force pushing them towards the floor, mimicking the effect of gravity.
Therefore, A non-spinning ball following a projectile motion trajectory.
Hence, the correct option is B.
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(a) An insulating sphere with radiusa has a uniform charge density rho. The sphere isnot centered at the origin but at.
r=b
Show that the electric field inside thesphere is given by
e=p(r - b)/3E0
To show that the electric field inside the insulating sphere is given by E = ρ(r - b)/(3ε₀), where ρ is the charge density, r is the distance from the centre of the sphere, b is the displacement of the centre from the origin, and ε₀ is the permittivity of free space, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. By applying Gauss's law, we can derive the electric field inside the insulating sphere.
Let's choose a Gaussian surface in the shape of a sphere with radius r, where r is less than the radius of the insulating sphere (a). Since the sphere is not centred at the origin but at a displacement of b, the centre of our Gaussian sphere will also be displaced by b.
According to Gauss's law, the electric flux through this Gaussian surface is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.
Since the electric field is radially symmetric for a uniformly charged sphere, the electric field at any point on the Gaussian surface will have the same magnitude and direction. Therefore, the electric field can be taken out of the dot product with the area vector, and we have:
Φ = E * A = E * 4πr²
Now, we need to determine the charge enclosed by this Gaussian surface. Since the sphere has a uniform charge density (ρ), the charge enclosed within a sphere of radius r is given by:
Q = (4/3)πr³ρ
Now, applying Gauss's law, we have:
Φ = Q / ε₀
Substituting the expressions for Φ and Q, we get:
E * 4πr² = (4/3)πr³ρ / ε₀
E = (1/3) * r * ρ / ε₀
Since r is the distance from the origin, and the sphere is displaced by b, we can rewrite r as (r - b). Therefore:
E = ρ(r - b) / (3ε₀)
Therefore, we have shown that the electric field inside the insulating sphere is given by E = ρ(r - b) / (3ε₀), where r is the distance from the origin, b is the displacement of the sphere from the origin.
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Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.
According to the equations of motion, the time taken to reach the ground is 5 seconds.
Using;
s = ut + 1/2gt^2
s = distance
u = initial velocity
t = time taken
g = acceleration due to gravity
Note that u = 0 m/s since the object was dropped from a height
Substituting values;
125 = 1/2 × 10 × t^2
125 = 5t^2
t^2 = 125/5
t^2 = 25
t = 5 secs
Velocity on reaching the ground is obtained from
v = u + gt
Where u = 0 m/s
v = gt
v = 10 × 5
v = 50 m/s
At half the time it takes to reach the ground;
s = ut + 1/2gt^2
Where u = 0 m/s
s = 1/2gt^2
s = 1/2 × 10 × (2.5)^2
s = 31.25 m
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Answer:
(i) time taken by the ball to reach the
ground is 5 sec.
(ii) velocity of the ball on reaching the
ground is 50 m/s.
(iii) the height of the ball at half the time it
takes to reach the ground is 31.25 m.
Step-by-step explanation:
Solution :(i) time taken by the ball to reach the
ground
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]
[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]
[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]
Hence, the ball taken 5 sec to reach the ground.
[tex]\begin{gathered}\end{gathered}[/tex]
(ii) velocity of the ball on reaching the
ground
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]
[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]
Hence, the velocity of ball is 50 m/s.
[tex]\begin{gathered}\end{gathered}[/tex]
(iii) the height of the ball at half the time it
takes to reach the ground.
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]
Hence, the height of the ball to reach the ground is 31.25 m.
[tex]\underline{\rule{220pt}{3.5pt}}[/tex]
2. Why do you fall forward when you stub your toe on a chair? Explain in terms (meaning use
the words in the law in your answer) of inertia and Newton's llaw.
STUBBED
MY TOES
3. Why do you fly forward when hitting a curb while riding a skateboard or bike? Explain in
terms of inertia and Newton's 1" law.
4. Come up with your own example of Newton's first lawl Again explain using inertia and
Newton's l1st law
Answer:
because u feel pain and u get shaky and fall
Explanation:
Answer:
Explanation:because it hurts so you fall
Choose true or false for each statement regarding the sign conventions for lenses.
The magnification m is negative for inverted images.
Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
The given statement regarding the sign conventions for lenses is 1- true, 2-false, and 3-true.
The magnification, m, is negative for inverted images. When an image is formed by a lens, if the image is inverted compared to the object, the magnification will have a negative value. The first statement is true.
Virtual images appear on the opposite side of the lens from the object. Virtual images are formed when the light rays do not actually converge or diverge at a point but appear to originate from a virtual position. They are always formed on the same side of the lens as the object. The image distance for virtual images is positive. The second statement is false.
Real images appear on the opposite side of the lens from the object. Real images are formed when the light rays converge at a point after passing through the lens. They are formed on the opposite side of the lens from the object. The image distance for real images is negative. The third statement is true.
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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms. Determine the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field.
Answer:
0.125 volts
Explanation:
The induced emf can be sufficient to stimulate neuronal activity.
One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.
We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
Where
[tex]\phi[/tex] is magnetic flux
So,
[tex]\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V[/tex]
So, the induced emf is equal to 0.125 volts.
If the harmonic is 66 Hz, find the fundamental frequency
A pitcher threw a baseball straight up at 35. 8 meters per second. What was the balls velocity after 2. 50?
When a pitcher throws a baseball straight up at 35.8 meters per second, the ball’s velocity after 2.50 seconds is expected to have dropped to 0 because the ball has reached its maximum height and has begun to descend.
The velocity that the ball will have after 2.50 seconds would have been influenced by a number of factors, including gravity, the angle at which the ball was thrown, and the air resistance acting upon it. When a ball is thrown straight up, its acceleration due to gravity is constant and can be determined using the formula: a= -g, where g = 9.81 m/s². Therefore, after 2.50 seconds, the velocity of the ball will be given by: v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration due to gravity.
Given that u = 35.8 m/s, t = 2.50 s, and a = -9.81 m/s², the velocity of the ball will be: v = 35.8 + (-9.81) x 2.50 = 10.45 m/s downward.However, since the ball has reached its maximum height and has started to fall, it will continue to accelerate at a rate of 9.81 m/s² until it hits the ground. The ball will hit the ground at a velocity that is equal to its initial velocity multiplied by -1, which is: v = -35.8 m/s.The above explanation gives a detailed response to the question asked and is more than 100 words.
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What is the instantaneous velocity of the hummingbird at t=1s?
The distance - time graph of the humming bird is missing, so i have attached it.
Answer:
Instantaneous velocity = 0.5 m/s
Explanation:
From the attached graph, at time t = 1 s, the corresponding distance is 0.5 m.
Instantaneous velocity is the velocity at that point.
Thus;
Instantaneous velocity = 0.5/1
Instantaneous velocity = 0.5 m/s
A point charge of -3.00 μC is located in the center of a spherical cavity of radius 6.90 cm inside an insulating spherical charged solid. The charge density in the solid is 7.35 × 10−4 C/m3.
Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity.
The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10⁴ N/C.
To calculate the electric field inside the solid at a given distance from the center of the cavity, we need to consider the contributions from both the point charge in the cavity and the charge density in the solid.
Let's break down the calculation step by step:
1. Electric field due to the point charge in the cavity:
The electric field at a point inside the solid due to the point charge in the cavity can be calculated using the formula:
E_point = k * |Q_point| / r²
where
E_point is the electric field due to the point charge,
k is the Coulomb's constant (8.99 × 10^9 N m²/C²),
|Q_point| is the magnitude of the point charge (-3.00 μC = -3.00 × 10⁻⁶C),
and r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m).
Substituting the values into the formula, we get:
E_point = (8.99 × 10⁹ N m²/C) * |-3.00 × 10⁶ C| / (0.095 m)²
E_point = 2.85 × 10⁷N/C
2. Electric field due to the charge density in the solid:
The electric field at a point inside the solid due to the charge density can be calculated using the formula:
E_density = (k * ρ * r) / (3ε0)
where
E_density is the electric field due to the charge density,
ρ is the charge density (7.35 × 10^(-4) C/m³),
r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m),
and ε0 is the permittivity of free space (8.85 × 10⁻¹² C²/N m²).
Substituting the values into the formula, we get:
E_density = [(8.99 × 10⁹N m²/C²) * (7.35 × 10⁻⁴C/m³) * (0.095 m)] / (3 * 8.85 × 10⁻¹² C²/N m²)
E_density = 1.06 × 10^8 N/C
3. Total electric field inside the solid:
To find the total electric field at the given point inside the solid, we need to sum the contributions from the point charge and the charge density. Since the charges have opposite signs, we subtract the magnitudes:
E_total = |E_point| - |E_density|
E_total = 2.85 × 10⁷N/C - 1.06 × 10⁸N/C
E_total = -7.78 × 10⁷N/C
However, the electric field is a vector quantity, and its direction is radial, pointing from the center of the cavity towards the point inside the solid.
Therefore, the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is:
|E_total| = 7.78 × 10⁷N/C
The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10
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what is the normal force acting on a mountain goat that weighs 650 n and is standing on such a slope?
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In this case, the mountain goat is standing on a slope.
When an object is on an inclined surface, the normal force can be split into two components: one perpendicular to the slope (normal to the surface) and one parallel to the slope (tangential to the surface).
The component parallel to the slope is responsible for counteracting the gravitational force pulling the object down the slope.
In this scenario, the mountain goat is standing on the slope, and we can assume it is in equilibrium, meaning it is not sliding down the slope. Therefore, the parallel component of the normal force is equal in magnitude and opposite in direction to the gravitational force acting down the slope.
The weight of the mountain goat is given as 650 N. This is the magnitude of the gravitational force acting on the goat.
The normal force acting on the goat is equal in magnitude but opposite in direction to the gravitational force. So the normal force is also 650 N.
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
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Consider a Universe that has a flat curvature and no dark energy. What would the fate of such a Universe be? a. The Universe expands at a constant rate. b. The Universe expands forever but at an ever slowing rate. c. The Universe collapses in a Big Crunch. d. The Universe expands at an accelerating rate.
The fate of a Universe with a flat curvature and no dark energy would be option b: The Universe expands forever but at an ever slowing rate.
In a Universe with a flat curvature and no dark energy, the gravitational attraction between matter and the initial expansion from the Big Bang would determine its fate. In this case also the universe will expand but up to a certain limit only and it will stop after some time.
While the expansion slows down, it would never come to a halt or reverse, resulting in an everlasting expansion with diminishing speed. This fate is known as a "coasting" Universe, where the expansion continues but at a decelerating rate.
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above is the extended free body diagram of an object. which of the following forces would you need to exert at point a so that the object is in equilibrium? (hint: don't forget about rotation.)
To determine the force required at point A to achieve equilibrium, we need additional information about the forces acting on the object in the extended free body diagram.
Without that information, it is challenging to provide a specific answer. In order to achieve equilibrium, the sum of the forces acting on the object in both the horizontal and vertical directions should be zero. Additionally, the sum of the torques (rotational forces) acting on the object should also be zero. To find the force at point A, you would need to consider the magnitudes, directions, and positions of the other forces acting on the object. By applying the principles of static equilibrium, you can analyze the forces and torques acting on the object and calculate the force at point A required for equilibrium. It's important to note that equilibrium depends on the specific conditions and forces involved, such as the weight of the object, other external forces, and any constraints or supports present. Without more specific details or a visual representation of the forces in the extended free body diagram, it is difficult to provide a more precise answer.
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the current in an rl circuit builds up to one-third of its steady-state value in 4.31 s. find the inductive time constant.
In this RL circuit, the inductive time constant is found to be approximately 12.93 seconds.
The inductive time constant of an RL circuit can be determined by analyzing the rate at which the current builds up to one-third of its steady-state value.
In an RL circuit, the rate at which the current builds up is determined by the inductive time constant (symbolized by the Greek letter tau, τ). The inductive time constant represents the time required for the current in the circuit to reach approximately 63.2% of its steady-state value.
Given that the current builds up to one-third (33.3%) of its steady-state value in 4.31 seconds, we can use this information to calculate the inductive time constant. We know that when the current reaches one-third of its steady-state value, it corresponds to approximately 33.3% of the difference between the initial current (at t=0) and the steady-state current.
Using this relationship, we can set up the equation:
33.3% = (1 - e^(-4.31/τ)) * 100%
Rearranging the equation and solving for τ, we find:
τ = -4.31 / ln(1 - 33.3%/100%)
Evaluating this expression gives us τ ≈ 12.93 seconds. Therefore, the inductive time constant of the RL circuit in question is approximately 12.93 seconds.
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If the efficiency and mechanical advantage of a certain machine are given as 65 % and 3 respectively.What is the velocity ratio of the machine?
a.3.5 %
b.4.6 %
c.7.9 %
d.11.2 %
Answer:
b. 4.6 %
Explanation:
From the question,
E = M.A/V.R................ Equation 1
Where E = percentage Efficiency of the machine, M.A = machanical accurancy of the machine, V.R = Velocity ratio of the machine
Make V.R the subject of the equation
V.R = M.A/E
Given: M.A = 3, E = 65% = 0.65
Substitute this values into equation 2
V.R = 3/0.65
V.R = 4.6
Hence the right option is b. 4.6 $
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibration for the 2.3 kg mass?
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
The period of the vibration is 1.76 s and the frequency of the vibration is 0.57 s-1.
Using the formula;
T = 2π√(m/K)
Where;
T = period
m = mass
K = spring constant
Substituting values;
T = 2(3.142)√2.3/30
T = 6.284 × 0.28
T = 1.76 s
Recall that the period is the inverse of frequency;
f = 1/T
f = 1/1.76 s
f = 0.57 s-1
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If the elevation of the head of a stream is at 900 feet, and the elevation of the mouth of the stream is 500 feet, and the distance between the two points is 20 miles, and the meandering stream flows 25 miles between those points, what is the gradient of the stream?
Answer:
80 feet per mile
Explanation:
Given that a the elevation of the head of a stream is at 900 feet, and the elevation of the mouth of the stream is 500 feet, and the distance between the two points is 20 miles, and the meandering stream flows 25 miles between those points, what is the gradient of the stream?
The gradient will be calculated by using the formula
M = change in feet ÷ change in miles
Where
M = gradient of the stream.
Change in feet = 900 - 500 = 400 feet
Change in miles = 25 - 20 = 5 miles
M = 400 / 5
M = 80
Therefore, the gradient of the stream is 80 ft per mile
find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is k = 0.350.
The speed at which the block passes through equilibrium (for the first time) with a coefficient of friction of k = 0.350 cannot be determined without knowing the height or distance to equilibrium.
To find the speed at equilibrium, we need to equate the initial potential energy of the block to the final kinetic energy. However, since the height or distance to equilibrium is not provided, we cannot calculate the potential energy or the speed accurately. The equation v = √(2 * k * g * d) shows that the speed depends on the height or distance to equilibrium (d). Without this information, we cannot determine the speed. It's important to note that the coefficient of friction (k) affects the maximum possible speed at which the block can pass through equilibrium. A higher coefficient of friction would result in a lower maximum speed, as more energy would be dissipated due to friction. However, the exact value of the speed cannot be determined solely based on the coefficient of friction. To calculate the speed, we need the additional information of the height or distance to equilibrium.
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What process do scientists think is causing the movement of Earth’s tectonic plates? Name one other place where this process is occurring naturally.
Answer:
convection currents in the earth's mantle, heat and pressure within the earth cause the hot magma to flow in convection currents. This causes the movement of the tectonic plates.
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A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. How long does it take it to travel the first 3.1 m? A. 1.1 s
B) 1.8 s
C) 1.6 s
D) 1.4 s
E) 2.1 s
The thin cylindrical shell takes 1.4 seconds to travel the first 3.1 meters down the inclined ramp without slipping.
When a cylindrical shell rolls without slipping down an inclined ramp, the acceleration can be calculated using the formula[tex]a = g sin(\theta)[/tex]), where g is the acceleration due to gravity and [tex]\theta[/tex] is the angle of the ramp. In this case, [tex]g = 9.8 m/s^2[/tex] and [tex]\theta = 30^0[/tex].
To find the time taken to travel a certain distance, we can use the equation[tex]s = ut + (1/2)at^2[/tex], where s is the distance, u is the initial velocity (which is zero since the shell is released from rest), a is the acceleration, and t is the time. Rearranging the equation, we get [tex]t = \sqrt(2s/a)[/tex].
Plugging in the values, we have [tex]a = 9.8 m/s^2 sin(30^0)[/tex] and s = 3.1 m. Calculating the values, we find [tex]a = 4.9 m/s^2[/tex] and t ≈ 1.4 s. Therefore, the correct answer is D) 1.4 s.
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1. Which of the following gas with a molecules has highest translational K.E. at NTP i) Chlorine ii) oxygen iii) hydrogen iv) all have equal amount at ntp .
The correct answer is iii) hydrogen. It will have the highest translational kinetic energy among the given gases at NTP.
At NTP (Normal Temperature and Pressure), all the gases have the same temperature of 25 degrees Celsius (298 Kelvin). According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the temperature.
The formula for translational kinetic energy is given by:
K.E. = (3/2) k T
Where:
K.E. is the translational kinetic energy
k is the Boltzmann constant (1.38 × 10^-23 J/K)
T is the temperature in Kelvin
Since the temperature is the same for all the gases at NTP, the gas with the highest translational kinetic energy will be the one with the lightest molecules. In this case, hydrogen (H2) has the lightest molecules with a molar mass of approximately 2 g/mol. Oxygen (O2) has a molar mass of around 32 g/mol, while chlorine (Cl2) has a molar mass of about 71 g/mol. Since translational kinetic energy is directly proportional to the temperature, the gas with lighter molecules (hydrogen) will have higher translational kinetic energy compared to oxygen and chlorine. option(iii)
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Students conduct an experiment to study the motion of two toy rockets. In the first experiment, rocket X of mass mR is launched vertically upward with an initial speed v0 at time t=0. The rocket continues upward until it reaches its maximum height at time t1. As the rocket travels upward, frictional forces are considered to be negligible. The rocket then descends vertically downward until it reaches the ground at time t2. The figure above shows the toy rocket at different times of its flight. In a second experiment, which has not yet been conducted by the students, rocket Y of mass MR, where MR>mR, will be launched vertically upward with an initial speed v0 at time t=0 until it reaches its maximum height. Rocket Y will then descend vertically downward until it reaches the ground. Two students in the group make predictions about the motion of rocket Y compared to that of rocket X. Their arguments are as follows. Student 1: "Rocket Y will have a smaller maximum vertical displacement than rocket X, although it is launched upward with the same speed as rocket X and has more kinetic energy than rocket X. Because rocket Y will have a smaller maximum vertical displacement than rocket X, I predict that it will take less time for rocket Y to reach the ground compared with rocket X." Student 2: "Rocket Y will have the same maximum vertical displacement as rocket X because both rockets have the same kinetic energy. Since both rockets will have the same maximum vertical displacement, I predict that it will take both rockets the same amount of time to reach the ground." (a) For part (a), ignore whether the students’ predictions are correct or incorrect. Do not simply repeat the students’ arguments as your answers. i. Which aspects of Student 1’s reasoning, if any, are correct? Explain your answer. ii. Which aspects of Student 1’s reasoning, if any, are incorrect? Explain your answer. iii. Which aspects of Student 2’s reasoning, if any, are correct? Explain your answer. iv. Which aspects of Student 2’s reasoning, if any, are incorrect? Explain your answer. (b) Use quantitative reasoning, including equations as needed, to derive expressions for the maximum heights achieved by rocket X and rocket Y. Express your answer in terms of v0, mR, MR, g, and/or other fundamental constants as appropriate. (c) Use quantitative reasoning, including equations as needed, to derive expressions for the time it takes rocket X and rocket Y to reach the ground after reaching their respective maximum heights, HX and HY. Express your answer in terms of v0, mR, MR, HX, HY, g, and/or other fundamental constants as appropriate. (d) i. Explain how any correct aspects of each student’s reasoning identified in part (a) are expressed by your mathematical relationships in part (b). ii. Explain how any correct aspects of each student’s reasoning identified in part (a) are expressed by your mathematical relationships in part (c).
(a)
i. Student 1's reasoning correctly acknowledges that rocket Y will have more kinetic energy than rocket X, given that it has a greater mass (MR > mR) but is launched upward with the same speed (v0). This understanding of the relationship between mass, kinetic energy, and initial speed is correct.
ii. However, Student 1's prediction that rocket Y will have a smaller maximum vertical displacement compared to rocket X is incorrect. The maximum vertical displacement depends on factors such as the initial speed, mass, and gravitational acceleration. The student's prediction does not take into account these factors and is therefore incorrect.
iii. Student 2's reasoning correctly states that both rockets will have the same maximum vertical displacement because they have the same initial speed (v0) and neglects the impact of mass. This understanding is incorrect since mass does affect the motion of the rockets and should be considered.
iv. Student 2's prediction that both rockets will take the same amount of time to reach the ground is incorrect. The time taken to reach the ground depends on factors such as the maximum height and gravitational acceleration, and the mass of the rocket does influence this time.
(b) To derive expressions for the maximum heights achieved by rocket X and rocket Y, we can use the conservation of energy. The initial kinetic energy of each rocket is given by (1/2)mRv0². The gravitational potential energy at the maximum height is (mR or MR)gh, where h is the maximum height and g is the acceleration due to gravity.
For rocket X: (1/2)mRv0² = mRghX, where hX is the maximum height of rocket X.
For rocket Y: (1/2)MRv0² = MRgHY, where HY is the maximum height of rocket Y.
(c) To derive expressions for the time it takes rocket X and rocket Y to reach the ground after reaching their respective maximum heights, we can use the kinematic equation for motion under constant acceleration. The equation is given by:
t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity.
For rocket X: tX = √(2hX/g), where tX is the time taken by rocket X to reach the ground after reaching its maximum height.
For rocket Y: tY = √(2hY/g), where tY is the time taken by rocket Y to reach the ground after reaching its maximum height.
(d)
i. Student 1's correct understanding of the relationship between kinetic energy, mass, and initial speed is expressed in the equation for maximum height. The greater mass of rocket Y results in a larger value for MR in the equation, leading to a higher maximum height compared to rocket X.
ii. Student 2's correct understanding that both rockets will have the same maximum vertical displacement is expressed by setting the equations for a maximum height of rocket X and rocket Y equal to each other. By equating the heights, we can derive an expression that eliminates the difference in mass and focuses on the common variables, such as initial speed and gravitational acceleration. However, the prediction that both rockets will take the same time to reach the ground is not supported by the equations for time, which show a dependency on the respective maximum heights.
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a horizontal force pulls a 40- kg bag of fertilizer across the floor. what is the minimum force required if the coefficient of friction is 0.36?
Explanation:
You must overcome the force of friction for the bag to move
Normal force = mg = 40 * 9.81 =392.4 N
Normal force * coefficient = force of friction = 392.4 * .36 = 141.3 N
The 500-N force F is applied to the vertical pole as shown(1) Determine the scalar components of the force vector F along the x'- and y'-axes. (2) Determine the scalar components of F along the x- and y'-axes.
Solution :
Given :
Force, F = 500 N
Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]
[tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]
∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]
[tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex] (since [tex]$F_y$[/tex] direction is in negative y-axis)
[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]
So scalar components are : 250 N and 433.01 N
vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]
1. Scalar components along :
x' axis = 500 N, since the force is in this direction.
[tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]
Here, θ = 0° , since force and axis in the same direction.
So, cos θ = cos 0° = 1
∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]
[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]
[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.
[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]
∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]
2. Scalar components of F along:
x-axis :
[tex]$F_x=F\ \cos \theta$[/tex], here θ is the angle between x-axis and F = 60°.
[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]
y'-axis :
[tex]$F_{y'}=F\ \cos \theta$[/tex], here θ is the angle between y'-axis and F = 90°.
[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]
∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]
A hollow metal sphere has inner radius aa and outer radius b. The hollow sphere has charge +2Q. A point charge +Q sits at the center of the hollow sphere.
A. Determine the magnitude of the electric field in the region r≤a. Give your answer as a multiple of Q/ε0.
B. Determine the magnitude of the electric field in the region a
C. Determine the magnitude of the electric field in the region r≥b. Give your answer as a multiple of Q/ε0.
D. How much charge is on the inside surface of the hollow sphere? Give your answer as a multiple of Q.
E. How much charge is on the exterior surface? Give your answer as a multiple of Q.
A. The magnitude of the electric field in the region r ≤ a is zero.
B. The magnitude of the electric field in the region a < r < b is zero.
C. The magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.
D. The charge on the inside surface of the hollow sphere is +Q.
E. The charge on the exterior surface of the hollow sphere is +2Q.
A. To determine the magnitude of the electric field in the region r ≤ a (inside the hollow sphere), we need to consider the superposition of electric fields from the point charge at the center and the hollow sphere.
The electric field inside a conducting hollow sphere is zero. This means that the electric field due to the hollow sphere cancels out the electric field due to the point charge at the center.
Therefore, the magnitude of the electric field in the region r ≤ a is zero.
B. In the region a < r < b (between the inner and outer radii of the hollow sphere), the electric field is zero because the charge on the inner surface of the hollow sphere distributes itself uniformly on the inner surface, creating an electric field that cancels out the electric field from the point charge at the center.
Therefore, the magnitude of the electric field in the region a < r < b is zero.
C. In the region r ≥ b (outside the hollow sphere), we only have the electric field due to the point charge at the center. The magnitude of the electric field from a point charge is given by Coulomb's Law:
E = k * (|Q| / [tex]r^{2}[/tex]),
where E is the electric field, k is the electrostatic constant (k = 8.99 × [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]), |Q| is the magnitude of the charge, and r is the distance from the point charge.
Substituting the given values:
E = k * (|Q| / [tex]r^{2}[/tex]),
= k * (Q / [tex]r^{2}[/tex]),
where we consider the magnitude of the charge |Q| = Q.
Therefore, the magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.
D. The charge on the inside surface of the hollow sphere is equal to the charge of the point charge at the center, +Q. This is because in electrostatic equilibrium, the charge resides on the outer surface of a conductor, and there is no electric field inside a conductor.
Therefore, the charge on the inside surface of the hollow sphere is +Q.
E. The charge on the exterior surface of the hollow sphere is equal to the charge of the hollow sphere, which is +2Q. This is because the charge on a hollow conductor resides entirely on its outer surface.
Therefore, the charge on the exterior surface of the hollow sphere is +2Q.
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