The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination.
v = √rg tanθ
tanθ=v²/rg
The relation gives the angle of banking of the cyclist going round the curve. Here v is the speed of the cyclist, r is the radius of the curve, and g is the acceleration due to gravity.
The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination. The normal force acting on the car while travelling through such a curving road has a horizontal component. The centripetal force needed to prevent skidding is provided by this component.
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Iodine-131 has a half-life of 8 days. How many grams of a 256 g sample would remain at the end of 56 days?
Answer:
Explanation:
The decay of a radioactive substance is governed by the formula:N(t) = N₀ e^(-λt)where N₀ is the initial amount of the substance, N(t) is the amount remaining after time t, and λ is the decay constant.The half-life of Iodine-131 is 8 days, which means that after each 8-day period, the amount remaining will be reduced by half. We can use this fact to calculate the amount remaining after 56 days.First, we need to find the decay constant λ, which is related to the half-life by the formula:λ = ln(2) / t½where ln(2) is the natural logarithm of 2, and t½ is the half-life.Substituting the values we have:λ = ln(2) / 8 days ≈ 0.08664 day^(-1)Next, we can use the formula for N(t) to calculate the amount remaining after 56 days:N(56) = N₀ e^(-λt) = 256 g e^(-0.08664 day^(-1) × 56 days) ≈ 22.6 gTherefore, approximately 22.6 grams of the original 256 gram sample would remain after 56 days.
12. Which of the following quantities, if any, is a
scalar quantity?
(A) A velocity of -2.0 m/s
(B) A potential energy of -2.0 J
(C) A momentum of -2.0 N.S
(D) A displacement of -2.0 m
(E) None of the above is a scalar quantity.
The correct answer is (B) A potential energy of -2.0 J.
A scalar quantity is a physical quantity that has magnitude only and no direction.
Potential energy is a scalar quantity because it is a measure of the energy stored in an object or system, which does not depend on direction or orientation.
Velocity (A) and displacement (D) are vector quantities, as they have both magnitude and direction. Momentum (C) is also a vector quantity, as it is the product of mass and velocity and therefore has both magnitude and direction.
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Which describes Hans Oersted’s work with electricity and magnetism?
He discovered electromagnetic induction after seeing a compass needle deflected by a flowing electric current.
He discovered electromagnetism after seeing a compass needle deflected by a flowing electric current.
He discovered electromagnetic induction after seeing a changing magnetic field generate an electric current. ( not C)
He discovered electromagnetism after seeing a changing magnetic field generate an electric current.
The correct statement is" He discovered electromagnetic induction after seeing a changing magnetic field generate an electric current."The correct option is C.
Hans Oersted was a physicist who conducted a number of experiments in the early 19th century that contributed to our understanding of the relationship between electricity and magnetism. In one of his most famous experiments, he discovered electromagnetic induction, which is the process by which a changing magnetic field generates an electric current in a nearby conductor.
Option A is incorrect because Oersted did not discover electromagnetic induction by seeing a compass needle deflected by a flowing electric current. Instead, he noticed that a compass needle was deflected when a current-carrying wire was brought near it.
Option B is incorrect because although Oersted observed a compass needle is deflected, it was not solely due to the flowing electric current. It was the interaction between the electric current and the magnetic field that caused the deflection.
Option D is incorrect because although a changing magnetic field can generate an electric current, Oersted's discovery was based on the observation of a current-carrying wire creating a magnetic field that affected a nearby compass needle.
Therefore, the correct option is C because Oersted discovered electromagnetic induction by seeing a changing magnetic field generate an electric current, which is the core concept of this phenomenon.
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State the difference between positive and negative zero error of a vernier calliper
(Please help it will mean a lot)
(
1. The graph above represents the nuclear decay of a radioactive element, measured using a radiation-detecting device. What is the half-life, in days, of this hypothetical element?
2. If the half-life of a given substance is 65 days, how long will it take for a
100-gram sample of the substance to decay until there is only 25 grams of the radioactive material remaining?
3. If a sample of radioactive isotopes takes 60 minutes to decay from 200 grams to
50 grams, what is the half-life of the isotope? Hint: First, determine how many times the sample has lost half of its mass, which tells you how many half-life cycles have occurred.
4. If a 500.0 g sample of technetium-99 decays to 62.5 g of technetium-99 remaining in 639,000 years, what is the half-life of technetium-99
1. The half-life is 2 days
2. The time taken is 130 days
3. The half-life is 30 minutes
4. The half-life is 213000 years
1. How do i determine the half-life?The half-life of a material is the time taken for half the material to decay.
From the above diagram, the following were obtained
Original amount = 80Half of original amount = 80 / 2 = 40Half-life =?From the diagram, we can see that the time taken to get to half of the element is 2 days.
Thus, we can conclude that the half-life is 2 days
2. How do i determine the time taken?First, we shall determine the number of half lives that has elapsed. This is obtained as follow:
Original amount (N₀) = 100 gAmount remaining (N) = 25 gNumber of half-lives (n) =?2ⁿ = N₀ / N
2ⁿ = 100 / 25
2ⁿ = 4
2ⁿ = 2²
n = 2
Finally, we shall determine the time taken. Details below
Half-life (t½) = 65 daysNumber of half-lives (n) = 2 Time taken (t) =?t = n × t½
t = 2 × 65
Time taken = 130 days
3. How do i determine the half-life?First, we shall determine the number of half lives that has elapsed. This is obtained as follow:
Original amount (N₀) = 200 gAmount remaining (N) = 50 gNumber of half-lives (n) =?2ⁿ = N₀ / N
2ⁿ = 200 / 50
2ⁿ = 4
2ⁿ = 2²
n = 2
Finally, we shall determine the half-life. Details below
Number of half-lives (n) = 2 Time taken (t) = 60 minutes Half-life (t½) = ?t½ = t / n
t½ = 60 / 2
Half-life = 30 minutes
4. How do i determine the half-life?First, we shall determine the number of half lives that has elapsed. This is obtained as follow:
Original amount (N₀) = 500 gAmount remaining (N) = 62.5 gNumber of half-lives (n) =?2ⁿ = N₀ / N
2ⁿ = 500 / 62.5
2ⁿ = 8
2ⁿ = 2³
n = 3
Finally, we shall determine the half-life. Details below
Number of half-lives (n) = 3 Time taken (t) = 639000 yearsHalf-life (t½) = ?t½ = t / n
t½ = 639000 / 3
Half-life = 213000 years
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What does the open universe theory say?
A. The universe is unchanging and will remain that way.
B. The mass of the universe is large enough for gravity to begin making it contract.
C. The universe may begin contracting due to gravity and lead to another big bang and continue this cycle over and over.
D. The mass of the universe is not large for its gravity to slow down the expansion, and it will continue indefinitely.
The open universe theory, states that the mass of the universe is not large enough for its gravity to slow down the expansion caused by the Big Bang, and it will continue expanding indefinitely.
What does the Open Universe Theory suggest?This theory also known as the unbounded or infinite universe theory suggests that the geometry of the universe is flat, meaning that parallel lines will never meet or intersect, and that the total energy of the universe is zero. In an open universe, the expansion will not stop, and galaxies will continue to move away from each other at an increasing rate. This theory is supported by observations of the cosmic microwave background radiation, which provide evidence for the early expansion of the universe.
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Suppose headphones were placed on a student, and a 500-Hz sound was fed to the left ear
slightly later than to the right ear. The student will discern the source of sound to be at about 45°
to the right of center. Why?
What is the approximate time delay in the response of the left ear?
Answer:
0.05 ms
Explanation:
The speed of sound in air is about 343 m/s. If the sound waves are coming from a source that is 1 meter away, then it will take about 3.43 microseconds for the sound waves to reach the right ear and about 3.53 microseconds for the sound waves to reach the left ear. This is a difference of about 0.097 microseconds, which is about 0.05 ms.
The human brain is very good at detecting small differences in time, and it uses this information to determine the direction of a sound source. When the sound waves reach the left ear slightly later than the right ear, the brain interprets this as a sound coming from the right side. The greater the difference in time, the further to the right the brain will perceive the sound to be.
In the case of a 500 Hz sound, the brain will perceive the sound to be coming from about 45° to the right of center if the sound waves reach the
Please answer this question I’ll give brainliest if it’s correct.
Q6. Saul (30 kg) is tobogganing down a hill on a toboggan that has a mass of 10 kg. Part way down, when he is still 20 m (measured vertically) above the bottom of the hill, he passes his sister Nadia at a velocity of 15 m/s. [9 marks]
(a) What is the total energy of Saul and the toboggan?
(b) What is the height of the hill?
(c) What will Saul’s velocity at the bottom of the hill be?
(d) What will Saul’s height be when he is moving at a velocity of 9.5 m/s?
Saul (30 kg) is tobogganing down a hill:
(a) total energy 14130 J.
(b) height of the hill 11.5 m.
(c) velocity 15.0 m/s.
(d) height 4.6 m.
How to calculate height and velocity?(a) The total energy of Saul and the toboggan is equal to the sum of their kinetic and potential energies. At the point when Saul passes his sister, he has kinetic energy due to his motion and potential energy due to his height above the bottom of the hill. The toboggan also has kinetic energy due to its motion. The total energy is given by:
Total energy = kinetic energy + potential energy
= (1/2)(mS + mT)v² + (mS + mT)gh
where mS = 30 kg is Saul's mass,
mT = 10 kg is the toboggan's mass,
v = 15 m/s is Saul's velocity,
g = 9.8 m/s² is the acceleration due to gravity, and
h = 20 m is the height above the bottom of the hill.
Plugging in the values:
Total energy = (1/2)(30 kg + 10 kg)(15 m/s)² + (30 kg + 10 kg)(9.8 m/s²)(20 m)
= 8250 J + 5880 J
= 14130 J
Therefore, the total energy of Saul and the toboggan is 14130 J.
(b) The height of the hill can be found by equating the initial potential energy at the top of the hill to the total energy at the point when Saul passes his sister. That is:
(mS + mT)gh = (1/2)(mS + mT)v² + (mS + mT)gh
Simplifying and solving for h:
h = (1/2)v²/g
= (1/2)(15 m/s)²/9.8 m/s²
= 11.5 m
Therefore, the height of the hill is 11.5 m.
(c) To find Saul's velocity at the bottom of the hill, use conservation of energy again. At the bottom of the hill, all the potential energy has been converted to kinetic energy. That is:
(mS + mT)gh = (1/2)(mS + mT)v²
Simplifying and solving for v:
v = √[2gh]
= √[2(9.8 m/s²)(11.5 m)]
= 15.0 m/s (to two significant figures)
Therefore, Saul's velocity at the bottom of the hill is 15.0 m/s.
(d) To find Saul's height when he is moving at a velocity of 9.5 m/s, we can use conservation of energy again. This time, equate the kinetic energy at this point to the initial potential energy. That is:
(1/2)(mS + mT)v² = (mS + mT)gh
Simplifying and solving for h:
h = v²/2g
= (9.5 m/s)²/(2)(9.8 m/s²)
= 4.6 m
Therefore, Saul's height when he is moving at a velocity of 9.5 m/s is 4.6 m.
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Example 9:
3.
Figure 5.24 shows a barrel of weight 1500 N
and radius 0.5 m that rests against a step of
height 0.2 m.
0.2 m
0.5 m
▲ Figure 5.24
(
What is the smallest horizontal force F th
the centre O needed to push the barrel over
the step
To drive the barrel over the step, the least horizontal force F necessary is 2366.16 N.
How to find horizontal force?To push the barrel over the step, the minimum force required should overcome the force of gravity acting on the barrel and the force of static friction between the barrel and the surface.
The perpendicular component of the weight is given as N = mgcosθ, where m = mass of the barrel,
g = acceleration due to gravity, and
θ = angle between the weight and normal to the surface.
In this case, θ as the inverse tangent of the ratio of the height of step to the distance from the edge of the step to the center of the barrel:
θ = tan⁻¹(0.2/0.5) = 0.39 radians
Therefore, the normal force is N = (1500)(9.81)cos(0.39) = 1443.6 N.
The force of static friction can be found as f = μsN,
where μs = coefficient of static friction.
Assume the coefficient of static friction between the barrel and the surface is 0.6.
f = (0.6)(1443.6) = 866.16 N.
The minimum force required to push the barrel over the step should overcome both these forces. Then, the smallest horizontal force that can push the barrel over the step is:
F = force of gravity + force of static friction
F = 1500 + 866.16
F = 2366.16 N.
Therefore, the smallest horizontal force F required to push the barrel over the step is 2366.16 N.
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14. Neglecting air resistance, what maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?
(a) 98 m
(b) 18 m
(c) 160 m
(d) 63 m
Answer:
D
Explanation:
The maximum height reached by a stone thrown straight up with an initial speed of 35 m/s can be found using the kinematic equation:
v^2f = v^2i - 2gh
where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (35 m/s, the magnitude of the velocity with which the stone is thrown upwards), g is the acceleration due to gravity (-9.8 m/s^2), and h is the maximum height reached by the stone.
Rearranging the equation, we get:
h = (vi^2)/(2g)
Substituting the given values, we have:
h = (35 m/s)^2 / (2 * 9.8 m/s^2)
= 62.6 m
Therefore, the maximum height reached by the stone is approximately 63 m.
The answer is (d).
Calculate the range in cm of 400keV beta ray(2.27 MeV from 90Y in bone of density 1.9g/cm3
The range of a 400 keV beta particle (2.27 MeV from 90Y) in bone with a density of 1.9 g/cm3 is approximately 0.000197 cm.
Using the Bethe-Bloch formula, the range of a 400 keV beta particle in bone with a density of 1.9 g/cm3 can be calculated as:
[tex]R = 0.415 * (E/ \rho Z^2) * (1/\beta^2 - 1)[/tex]
where R is the range in cm, E is the kinetic energy of the beta particle in MeV, ρ is the density of the material in g/cm3, Z is the atomic number of the material, and β is the velocity of the beta particle as a fraction of the speed of light.
Plugging in the given values, we get:
[tex]R = 0.415 * (0.4 / (1.9 * 22.5^2)) * (1 / (1 - (0.4 / 2.27)^2)) \\R = 0.415 * (0.4 / 966.375) * (1 / 0.858) \\R = 0.000197 cm[/tex]
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energy transferred from one thing to another when the gulf balls collide?
This is based on the principle of conservation of energy and momentum.
When the collision of golf balls takes place the energy gets transferred from one ball to the other.
The golf balls experience a force that causes them to change their form and also direction. During the collision, the mechanical energy is converted into heat, sound, and other forms of energy. The rest of energy is used to move in a new direction.
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A displacement vector is 23 km in length and directed 65° south of east. What are the components of this vector?
Eastward
Component Southward. Component
(a) 21 km 9.7 km
(b) 23 km 23 km
(c) 23 km 0 km
(d) 9.7 km 21 km
If a displacement vector is 23 km in length and directed 65° south of east. the components of this vector is: (a) 21 km 9.7 km.
What is the components of this vector?The displacement vector can be resolved into its eastward and southward components using trigonometry. Let's call the eastward component "x" and the southward component "y".
From the given information, we know that the displacement vector makes an angle of 65° south of east. This means that the angle between the vector and the eastward axis is 90° - 65° = 25°.
Using trigonometry, we can relate the length of the vector to its components:
cos 25° = x / 23
sin 25° = y / 23
Solving for x and y, we get:
x = 23 cos 25° ≈ 21 km
y = 23 sin 25° ≈ 9.7 km
Therefore, the answer is (a) 21 km eastward component and 9.7 km southward component.
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OK, once again we have a pendulum, this time of length 1.06 m, which you release from rest at an angle of 41.2 degrees to the vertical. What will be the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical?
The speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical is 3.02 m/s.
A pendulum is a weight suspended from a fixed point that swings back and forth due to the force of gravity.
Based on the given information, we have a pendulum of length 1.06 m and it is released from rest at an angle of 41.2 degrees to the vertical. We need to find the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical.
To solve this problem, we can use the conservation of mechanical energy. At the highest point of the pendulum's swing, all of its energy is in the form of potential energy, and at the lowest point of its swing, all of its energy is in the form of kinetic energy. Therefore, we can write:
PE_max = KE_min
where PE_max is the potential energy at the maximum height and KE_min is the kinetic energy at the lowest point.
The potential energy of a pendulum is given by:
PE = mgh
where m is the mass of the pendulum, g is the acceleration due to gravity, and h is the height above some reference point.
The kinetic energy of a pendulum is given by:
KE = (1/2)mv^2
where v is the speed of the pendulum.
First, we need to find the vertical height difference between the pendulum's highest and lowest points. To do this, we can use trigonometry:
h = L(1 - cosθ)
where L is the length of the pendulum and θ is the initial angle to the vertical. Substituting the given values, we get:
h = 1.06(1 - cos(41.2°)) = 0.654 m
Next, we can use the conservation of mechanical energy to find the speed of the pendulum at the lowest point of its swing. At this point, all of the potential energy has been converted into kinetic energy, so we can write:
PE_max = KE_min
mgh = (1/2)mv^2
Canceling out the mass, we get:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
where g is the acceleration due to gravity. Substituting the given values, we get:
v = sqrt(2(9.81 m/s^2)(0.654 m)) = 3.78 m/s
Finally, we need to find the speed of the pendulum when it reaches an angle of 20.6 degrees above the vertical. At this point, the pendulum has a potential energy of:
PE = mgh' = mgh cos(20.6°)
where h' is the height of the pendulum at this point. To find h', we can use trigonometry:
h' = L(1 - cosθ')
where θ' is the angle above the vertical. Substituting the given values, we get:
h' = 1.06(1 - cos(20.6°)) = 0.242 m
Substituting the values for h' and solving for the kinetic energy, we get:
KE = PE_max - mgh' = mgh - mgh'
Substituting the known values, we get:
KE = (1 kg)(9.81 m/s^2)(0.654 m) - (1 kg)(9.81 m/s^2)(0.242 m) = 5.11 J
Now, we can solve for the speed at this point:
KE = (1/2)mv^2
5.11 J = (1/2)(1 kg)v^2
v = sqrt((2)(5.11 J)/(1 kg)) = 3.02 m/s
Therefore, The pendulum is moving at a speed of 3.02 m/s when it reaches an angle of 20.6 degrees above vertical.
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How does our viewing angle from earth affect what type of agn we observe?
Our viewing angle from Earth can affect the type of AGN (Active Galactic Nuclei) we observe. If we view an AGN from the direction of its accretion disk, we see a quasar, which is an incredibly bright object.
How does the viewing angle from earth affect how we observe?if we view the same AGN from a different angle, we may see a less bright active galactic nucleus, such as a Seyfert galaxy or a blazar. This is because the orientation of the accretion disk and the angle of the jets relative to our line of sight can affect the amount of radiation that reaches us.
In some cases, the orientation of the AGN can also affect the appearance of its emission lines. For example, in a type 2 Seyfert galaxy, which is viewed at an angle where the accretion disk is obscured by a thick torus of dust and gas, the broad emission lines that are typically seen in type 1 Seyfert galaxies may be absent or significantly narrower. Thus, the viewing angle from Earth plays an important role in the study of AGN and their properties.
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22. How do we correct the issue of flipped imagery caused by mirrors?
To correct the issue of flipped imagery caused by mirrors you can use a technique called "mirror flipping". Mirror flipping involves using a second mirror to reflect the reflected image from the first mirror which then flips it back to its original orientation
Alternatively you can use a prism to correct the orientation of the image. A prism is a transparent object that can bend light. By placing it infront of the mirror you can reflect the twice, effectively helping in the correction of the orientation of the image
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describe the energy changes in a mass spring system that is oscillating horizontally explain how this changes of the system is vibrating vertically
Answer:
In a horizontal mass spring system, the energy changes between potential energy and kinetic energy. When the spring is at its equilibrium position, the mass has potential energy stored in the spring, and no kinetic energy. As the mass is displaced from its equilibrium position and begins to move, potential energy is converted into kinetic energy. The maximum kinetic energy occurs when the mass is at the maximum displacement.
As the mass moves back towards the equilibrium position, the kinetic energy is converted back into potential energy stored in the spring. The maximum potential energy is reached when the mass reaches the equilibrium position. The energy then changes back to kinetic energy as the mass moves past the equilibrium position again.
If the same mass spring system is vibrating vertically instead of horizontally, gravitational potential energy also comes into play. When the mass is at its highest point, it has maximum gravitational potential energy and minimum kinetic energy. As it falls towards the equilibrium position, the potential energy is converted into kinetic energy. At the equilibrium position, the kinetic energy is maximum and potential energy is minimum. As the mass moves back up towards the maximum point, the kinetic energy is converted back into gravitational potential energy. This process continues as the system oscillates vertically.
ray of light exits from a metal with a refractive index of 1.75 travelling to the air the angle of refraction is 25°. what is the angle of deviation
Answer:
Explanation:
To find the angle of deviation, we can use the formula:
angle of deviation = (refractive index of metal - refractive index of air) x angle of incidence
The angle of incidence can be calculated using Snell's law:
sin(angle of incidence) / sin(angle of refraction) = refractive index of air / refractive index of metal
sin(angle of incidence) / sin(25°) = 1 / 1.75
sin(angle of incidence) = 0.5714
angle of incidence = 34°
Now we can substitute this value into the formula for angle of deviation:
angle of deviation = (1.75 - 1) x 34°
angle of deviation = 21°
Therefore, the angle of deviation is 21°.
What type of atomic radiation will most deeply penetrate matter?
Multiple Choice
Beta radiation
Gamma radiation
Alpha radiation
What happens as a protostar contracts?
A. More hydrogen is produced to become fuel for the star.
B. The temperature rises.
C. The hydrogen fuses into iron.
D. The temperature drops.
Answer:
As a protostar contract, the temperature rises. This increase in temperature leads to the initiation of nuclear fusion reactions, where hydrogen atoms fuse together to form helium, releasing energy in the process. This energy causes the protostar to heat up and begin to emit light, eventually becoming a stable star.
The answer is B. The temperature rises.
A car starting from rest has an acceleration 0.5m/s after 1 minute then what will be the final velocity of the car
Answer:
The final velocity is 30 m/s
Explanation:
Use the formula:
[tex]V_{f} = V_{i} +a*t\\V_{f} = 0 + 0.5*60\\V_{f} = 30 m/s[/tex]
A simple pendulum consisting of a bob of mass m
attached to a string of length L
swings with a period T
.
Part A
If the bob's mass is doubled, approximately what will the pendulum's new period be?
Part B
If the pendulum is brought on the moon where the gravitational acceleration is about g/6
, approximately what will its period now be?
T/6
T/√6
√6T
6T
Part C
If the pendulum is taken into the orbiting space station what will happen to the bob?
View Available Hint(s)for Part C
If the pendulum is taken into the orbiting space station what will happen to the bob?
It will continue to oscillate in a vertical plane with the same period.
It will no longer oscillate because there is no gravity in space.
It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
It will oscillate much faster with a period that approaches zero.
Answer: Part A:
The period of a simple pendulum is given by the formula: T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. If the mass of the bob is doubled, the period of the pendulum will change. To see how much it changes, we can use the fact that the period depends only on the length of the pendulum and the acceleration due to gravity, and not on the mass of the bob. Therefore, the new period of the pendulum will be the same as the old period: T.
Part B:
If the pendulum is brought to the moon, where the acceleration due to gravity is about g/6, the new period of the pendulum can be found using the same formula as before: T = 2π√(L/g). However, now we need to use the value of the acceleration due to gravity on the moon, which is g/6. Therefore, the new period of the pendulum is T' = 2π√(L/(g/6)) = 2π√(6L/g) = √6T.
Therefore, the answer is: T/√6.
Part C:
If the pendulum is taken into an orbiting space station, the bob will continue to oscillate in a vertical plane with the same period as it did on the surface of the Earth. This is because the period of the pendulum depends only on the length of the pendulum and the acceleration due to gravity, and not on the location of the pendulum. In the space station, both the pendulum and the point to which it is attached are in free fall, but they are falling together and maintaining their relative positions, so the pendulum will continue to oscillate as before.
a student rises their 15 kg back pack from the froor at a constant velocity of 5.0 m/s. How much force must the student apply
a student rises their 15 kg back pack from the floor at a constant velocity of 5.0 m/s then they apply of 37.5 N of force.
Force is responsible for the motion of an object. it produces acceleration in the body. According to newton's second law force is mass times acceleration i.e. F =ma. Its SI unit is N which is equivalent to kg.m/s². There are two types of forces, balanced force and unbalanced force.
In this problem student rises 15 kg of backpack at constant velocity and it happens in 2s, so consider time t = 2s.
Force is change in velocity with respect to time multiplied by mass.
F = m dv/dt
F = 15 kg × 5.0 m/s/2s
F = 37.5 N
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A wave oscillates 5.0 times a second and has a speed of 4.0 m/s
A.What is the frequency of this wave?
Express your answer to two significant figures and include the appropriate units.
b.What is the period of this wave?
Express your answer to two significant figures and include the appropriate units.
c.What is the wavelength of this wave?
Express your answer to two significant figures and include the appropriate units.
may or may not be lying. \(^_^)/
a. f = 5.0 Hz
b. T = 1/f
1/5.0 Hz
period of 0.20 s
c. λ = v/f
4.0 m/s / 5.0 Hz
wavelength of 0.80 m
IG:whis.sama_ent
A wave oscillates 5.0 times a second and has a speed of 4.0 m/s.
A. The frequency of any wave is the number of times it oscillates in one second. That's given in the question.
Frequency = 5.0 Hertz.
B. The period of any wave is the reciprocal of its frequency.
Period = (1 / 5.0 Hz)
Period = 0.2 second .
C. The wavelength of any wave is (speed / frequency).
Both of these numbers are given in the question.
Wavelength = (4.0 m/s) / (5.0 Hz) .
Wavelength = 0.8 meter .
Please help me in this question
Answer:
A. Yes, the force of the engine does work on the car as it makes the car move forward.
B. Yes, the kinetic energy of the car increases as the car gains speed due to the force of the engine.
C. The gravitational potential energy of the car remains constant as the car is moving on a horizontal road, and there is no change in its height.
D. Yes, the tractive force of the engine changes the mechanical energy of the car as it is a type of external force that transfers energy to the car and increases its kinetic energy.
Explanation:
A 25.00-V battery is used to supply current to a Ω10-kΩ resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current through the resistor? (b) What is the power dissipated by the resistor? (c) What is the power input from the battery, assuming all the electrical power is dissipated by the resistor? (d) What happens to the energy dissipated by the resistor?
a. The current through the resistor is 2.50 milliamperes. b. The power dissipated by the resistor is 62.5 milliwatts. c. The power input from the battery is also 62.5 milliwatts. d. The energy dissipated by the resistor is converted into heat energy and is lost to the surroundings.
The current through the resistor can be found using Ohm's law:
I = V/R = 25.00 V / 10 kΩ = 2.50 mA
The power dissipated by the resistor can be found using the formula:
P = I² R
Substituting the values, we get:
P = (2.50 x 10⁻³ A)² x 10 kΩ = 62.5 mW
The power input from the battery can be found using the formula:
P = IV
Substituting the values, we get:
P = (2.50 x 10⁻³ A) x 25.00 V = 62.5 mW
The energy dissipated by the resistor is converted into heat energy and is lost to the surroundings. This is known as joule heating or ohmic heating, where the electrical energy is converted into thermal energy due to the resistance of the material.
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The model of a ceiling fan shown in the figure consists of a uniform solid cylinder, of radius R = 0.067 m and mass MC = 1.8 kg, and two long uniform rods, each of length L = 0.94 m and mass MR = 3.4 kg, that are attached to the cylinder and extend from its center. Ignore the vertical rod that connects the fan to the motor.
(a) Enter an expression, in terms of the quantities defined in the problem, for the moment of inertia of each rod about the rotation axis.
(b) Enter an expression, in terms of the quantities defined in the problem, for the moment of inertia of the cylinder about the rotation axis.
(c) Enter an expression, in terms of the quantities defined in the problem, for the moment of inertia of the whole fan about the rotation axis.
(d) Calculate the moment of inertia, in units of kilogram meters squared, of the whole fan about the rotation axis.
a) The moment of inertia of each rod about the rotation axis is MR*L2/12.
What is axis?Axis is a reference line used for plotting points on a graph. It is used to indicate a direction and to measure distances. In a two-dimensional graph, the x-axis is typically used to represent the horizontal axis, while the y-axis is used to represent the vertical axis. In a three-dimensional graph, an additional z-axis is used to represent the third dimension. On the x-axis, the points are typically labeled with the independent variable, while the y-axis is labeled with the dependent variable. By plotting points on a graph, relationships between two or more variables can be seen and analyzed. This can be used to identify trends and make predictions about data.
b) The moment of inertia of the cylinder about the rotation axis is MC*R2.
c) The moment of inertia of the whole fan about the rotation axis is the sum of the moment of inertia of the rods and the cylinder, so it is MR*L2/12 + MC*R2.
d) Substituting in the values from the problem, the moment of inertia of the whole fan about the rotation axis is 3.4*0.942/12 + 1.8*0.0672 = 0.4086 kg m2.
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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 51.0 cm
. The explorer finds that the pendulum completes 110 full swing cycles in a time of 133 s
.
Part A
What is the magnitude of the gravitational acceleration on this planet?
Express your answer in meters per second per second.
The period of a simple pendulum is given by the formula:
T = 2π√(L/g)
where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.
We can rearrange this formula to solve for g:
g = (4π²L)/T²
Plugging in the given values:
L = 51.0 cm = 0.510 m
T = 133 s / 110 = 1.209 s
(Note that we divide the total time by the number of cycles to get the time for one cycle.)
So,
g = (4π² × 0.510 m) / (1.209 s)²
= 9.57 m/s²
Therefore, the magnitude of the gravitational acceleration on this planet is approximately 9.57 m/s².
*IG:whis.sama_ent*
Answer: 7.68 m/s²
Explanation: The period of a simple pendulum is given by the formula:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.
In this problem, the pendulum completes 110 full swing cycles in a time of 133 s, which means that the period of the pendulum is:
T = 133 s / 110 = 1.209 s
Substituting this into the formula above and solving for g, we get:
g = (4π²L) / T² = (4π²)(0.51 m) / (1.209 s)² ≈ 7.68 m/s²
Therefore, the magnitude of the gravitational acceleration on this planet is approximately 7.68 m/s².
A uniform horizontal rod of mass 1.5 kg and length 1.9 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by
m.02
I=
12
930.
-pivot
1.5 kg
1.9 m -
If a 1.5 N force at an angle of 95° to the hot-izontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9.8 m/s?
Answer in units of rad/s?
The magnitude of the resulting angular acceleration about the pivot point is approximately 0.0021 rad/s^2.
Magnitude of the resulting angular accelerationTo solve this problem, we need to use the equation for the torque on a rigid body:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We can also use the fact that the torque is given by:
τ = r × F
where r is the vector from the pivot point to the point of application of the force, and F is the force. The vector cross product r × F gives the torque about the pivot point.
First, we need to find the vector r. Since the force is acting at an angle of 95° to the horizontal, the vertical component of the force is given by:
Fy = F sin(95°) = 1.5 sin(95°) ≈ 0.15 N
The horizontal component of the force is given by:
Fx = F cos(95°) = 1.5 cos(95°) ≈ 0.04 N
The vector r points vertically downward from the pivot point and has a magnitude equal to half the length of the rod:
|r| = 0.5(1.9 m) = 0.95 m
Therefore, the vector r is given by:
r = -0.95j
where j is the unit vector in the vertical direction.
The torque about the pivot point is given by:
τ = r × F = (-0.95j) × (0.04i + 0.15j) = -0.15k
where i is the unit vector in the horizontal direction and k is the unit vector in the direction perpendicular to the plane of the rod and pointing out of the page.
Substituting the moment of inertia and torque into the equation for angular acceleration, we have:
α = τ/I = (-0.15k)/(930.02/12) ≈ -0.0021 rad/s^2
Therefore, the magnitude of the resulting angular acceleration about the pivot point is approximately 0.0021 rad/s^2.
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A car is traveling at a speed of 30m/s and leaves the ramp at a 37 degree angle. What is the total hang time of the car?
Explanation:
INITIAL Vertical velocity is given by
30 m/s * sin 37
then gravity begins to slow it down
30 sin37 - (9.81) t = vertical velocity at t
when v = 0 , the car is at its apex and will fall back down in the same amount of time
0 = 30 sin37 - 9.81 t shows t = 1.84 seconds to peak
then another 1.84 seconds to fall to the ground total = 3.7 seconds