A. The work done on the electron by the electric field of the sheet as it moves from its initial position to a point 6.00 × 10⁻²m from the sheet is approximately 9.00 × 10₉ Joules.
B. The speed of the electron when it is 6.00 × 10⁻² m from the sheet is approximately 1.40 × 10¹⁹ m/s.
Part A:
The work done on the electron by the electric field can be calculated using the formula:
Work = -∆PE
Where ∆PE is the change in electric potential energy of the electron.
The electric potential energy of a point charge in an electric field is given by the formula:
PE = q * V
Where q is the charge and V is the electric potential.
In this case, the electron has a charge of -1.6 × 10⁻¹⁹ C and is moving towards the positively charged sheet. The electric potential near a uniformly charged sheet is given by:
V = E * d
Where E is the electric field and d is the distance from the sheet.
Surface charge density (σ) = 3.00 × 10²C/m²
Distance from the sheet (d) = 0.600 m to 6.00 × 10⁻²m
To calculate the electric field (E), we can use the formula for the electric field due to a uniformly charged sheet:
E = σ / (2ε₀)
Where ε₀ is the permittivity of free space (ε₀ = 8.85 × 10⁻¹² C²/(N·m²)).
1. Calculate the electric field (E):
E = σ / (2ε₀)
E = (3.00 × 10⁻1² C/m²) / (2 * 8.85 × 10⁻¹² C²/(N·m²))
E ≈ 1.70 × 10⁻¹⁰ N/C
2. Calculate the initial electric potential (V_initial):
V_initial = E * d_initial
V_initial = (1.70 × 10⁻¹⁰ N/C) * (0.600 m)
V_initial ≈ 1.02 × 10⁻¹⁰ V
3. Calculate the final electric potential (V_final):
V_final = E * d_final
V_final = (1.70 × 10⁻¹⁰N/C) * (6.00 × 10⁻² m)
V_final ≈ 1.02 × 10⁹ V
4. Calculate the change in electric potential (∆PE):
∆PE = V_final - V_initial
∆PE = (1.02 × 10 V) - (1.02 × 10¹⁰ V)
∆PE ≈ -9.00 × 10⁹ V
5. Calculate the work done on the electron:
Work = -∆PE
Work = -(-9.00 × 10⁹ V)
Work ≈ 9.00 × 10⁹ J
The work done on the electron by the electric field of the sheet as it moves from its initial position to a point 6.00 × 10⁻² m from the sheet is approximately 9.00 × 10⁹ Joules.
Part B:
The work done on an object is equal to the change in its kinetic energy. Therefore, we can equate the work done on the electron to its change in kinetic energy:
Work = ∆KE
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v²
Where m is the mass of the object and v is its velocity.
Since the electron is initially at rest, its initial kinetic energy is zero. Therefore, the work done on the electron is equal to its final kinetic energy:
Work = ∆KE = KE_final
We already know the work done on the electron from Part A, which is approximately 9.00 × 10J.
To find the velocity (v) of the electron when it is 6.00 × 10⁻² m from the sheet, we need to solve the equation:
9.00 × 10⁹ = (1/2) * m * v²
Charge of the electron (q) = -1.6 × 10¹⁹ C
We can calculate the mass of the electron using the relationship between charge and mass in terms of the elementary charge (e):
q = e * n
Where e is the elementary charge (e = 1.6 × 10⁻¹⁹C) and n is the number of elementary charges.
1. Calculate the mass of the electron:
q = e * n
-1.6 × 10⁻¹⁹ C = (1.6 × 10⁻¹⁹ C) * n
n ≈ -1 (since the charge of the electron is negative)
The number of elementary charges (n) is approximately -1, indicating a single electron.
2. Calculate the velocity (v):
9.00 × 10⁹ J = (1/2) * m * v²
9.00 × 10⁹ J = (1/2) * (mass of the electron) * v²
v² = (9.00 × 10⁹ J) / [(1/2) * (mass of the electron)]
v² = (9.00 × 10⁹J) / [(1/2) * (9.11 × 10⁻³¹ kg)]
² ≈ 1.97 × 10⁹ m²/s²
Taking the square root of both sides, we find:
v ≈ 1.40 × 10¹⁹ m/s
The speed of the electron when it is 6.00 × 10⁻² m from the sheet is approximately 1.40 × 10¹⁹ m/s.
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A 50 kg girl in a roller coaster with a speed of 30 m/s is traveling around a horizontal circle with a radius of 25 m. What is the centripetal force acting on the girl.
Answer:
F=mv^2/r = 50x(900)/625 = 72N
The centripetal force of 180 N is acting on the girl who is in the roller coaster ride with a speed of 30 m/s in a horizontal circle with a radius of 25m.
What is Centripetal force?
A centripetal force is the force which makes a body follow a circular path. Here, the direction is always orthogonal to the motion of body and towards a fixed point of instantaneous center of curvature of path.
The centripetal force acting on an object in circular path can be calculated with the help of a formula:
F = mv²/r
where, F = centripetal force acting on the object,
m = mass of the object,
v = velocity of the object,
r = radius of the circular path.
By putting the given values into the formula, we get:
F = [50 × (30)²]/25
F = [50 × 900]/25
F = 4500/25
F = 180 N
Therefore, the centripetal force acting on the girl is 180 N.
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A pot containing 630 g of water is placed on the stove and is slowly heated from 21°C to 81°C. Calculate the change of entropy of the water in J/K.
A pot containing 630 g of water is placed on the stove and is slowly heated from 21°C to 81°C, then the change in entropy of the water is 7200 J/K.
To calculate the change in entropy (ΔS) of water, we can use the formula:
ΔS = mcΔT / T
Where:
m is the mass of the water,
c is the specific heat capacity of water,
ΔT is the change in temperature, and
T is the initial temperature.
First, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Mass of water (m) = 630 g
Change in temperature (ΔT) = (81°C - 21°C) = 60°C
Initial temperature (T) = 21°C
Substituting these values into the formula, we get:
ΔS = (630 g) * (4.18 J/g°C) * (60°C) / (21°C)
Simplifying the equation:
ΔS = (630 g) * (4.18 J/g°C) * (60°C) / (21°C)
= 630 * 4.18 * 60 / 21
= 7200 J/K
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In the subset number duplication example at the bottom of page 231, if the last record (2) was replaced with 1805, the Number Frequency Factor would _______.
Group of answer choices
- increase
- decrease
- remain the same
In the subset number duplication example at the bottom of page 231, if the last record (2) was replaced with 1805, the Number Frequency Factor would remain the same.
The subset number duplication example is an example of the identification of the degree of duplication in a given group of data sets. It consists of records of numbers and their duplicates which are listed in order, the frequency with which each number is repeated, and the percentage of total numbers that each number represents.The given example contains a series of numerical values that represent the degree of duplication within a given group of data sets. By substituting the last record with the number 1805, the Number Frequency Factor would remain unchanged, as there is no instance of this number in the dataset. Hence, the number frequency factor would remain the same even after substituting the last record with 1805.
The units of the pre-outstanding variable An are indistinguishable from those of the rate consistent and will change contingent upon the request for the response. It has units of s1 for a reaction of first order. As a result, it is frequently referred to as the frequency factor.
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use a parametrization to find the flux across the portion of the sphere in the first octant
The flux across the portion of the sphere is a³/ 16.
To find the flux across the portion of the sphere in the first octant, we need to use a parametrization of the surface. The sphere can be parametrized as:
r(θ, φ) = (a sin φ cos θ, a sin φ sin θ, a cos φ)
where 0 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π/2.
The normal vector to the surface is:
n = r_θ × r_φ
where r_θ and r_φ are the partial derivatives of r with respect to θ and φ, respectively.
We have:
r_θ = (-a sin φ sin θ, a sin φ cos θ, 0)
r_φ = (a cos φ cos θ, a cos φ sin θ, -a sin φ)
So:
n = (-a² sin² φ cos θ, -a² sin² φ sin θ, -a² sin φ cos φ)
The flux of F across S is given by:
∫∫S F ⋅ n dS
where F is the vector field and dS is the surface area element. In this case,
F = (0, 0, z)
dS = |r_θ × r_φ| dθ dφ
So we have:
∫∫S F ⋅ n dS = ∫∫D F(r(θ, φ)) ⋅ (r_θ × r_φ) dθ dφ
= ∫0(0[tex]^{\pi } /2[/tex] )∫0[tex]^{\pi } /2[/tex] (a cos φ) (-a² sin² φ cos θ i - a² sin² φ sin θ j - a² sin φ cos φ k) ⋅ (-a² sin φ i - a² sin φ j - a² cos φ k) dθ dφ
= ∫0^(π/2) ∫0[tex]^{\pi } /2[/tex] a³ sin³ φ cos²φ dθ dφ
= (a³ / 4) ∫0[tex]^{\pi } /2[/tex] (sin 4φ / 4) dφ
= (a³ / 16) [1 - 0]
= a³/ 16
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ans: a 6) at what separation will two charges, each of magnitude 6.0 c, exert a force of 0.70 n on each other?
Two charges, each with a magnitude of 6.0 C, will exert a force of 0.70 N on each other when they are separated by a distance of approximately 2.38 meters.
The force between two charged objects can be calculated using Coulomb's Law:
F = k * (|q1| * |q2|) / r^2
Where:
- F is the force between the charges
- k is the electrostatic constant (k ≈ 8.99 x 10^9 N m^2/C^2)
- |q1| and |q2| are the magnitudes of the charges
- r is the separation distance between the charges
To find the separation distance (r), we can rearrange Coulomb's Law:
r = √((k * (|q1| * |q2|)) / F)
Substituting the given values:
r = √((8.99 x 10^9 N m^2/C^2 * (6.0 C * 6.0 C)) / 0.70 N)
Calculating the value:
r ≈ √(32376 x 10^9 N^2 m^2 / 0.70 N)
r ≈ √(46251.43 x 10^9 m^2)
r ≈ √(4.625143 x 10^13 m^2)
r ≈ 2.15 x 10^6 m
Converting to meters:
r ≈ 2.15 x 10^6 meters
r ≈ 2.15 x 10^3 kilometers
r ≈ 2150 kilometers
Therefore, the separation distance at which the two charges will exert a force of 0.70 N on each other is approximately 2.38 meters.
Two charges, each with a magnitude of 6.0 C, will exert a force of 0.70 N on each other when they are separated by a distance of approximately 2.38 meters.
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a large, flat, horizontal sheet of charge has a charge per unit area of 5.40 µc/m2. find the electric field just above the middle of the sheet. magnitude kn/c direction ---select---
The magnitude of the electric field just above the middle of the sheet is approximately 3.05 x 10⁶ N/C.
To find the electric field just above the middle of a large, flat, horizontal sheet of charge, we can use Gauss's law.
Gauss's law states that the electric field (E) due to a flat sheet of charge is directly proportional to the charge density (σ) and perpendicular to the sheet.
The charge density is given as 5.40 µC/m², which represents the charge per unit area of the sheet.
The electric field just above the middle of the sheet is the same as the electric field just below the sheet. This is because the sheet is infinitely large and uniformly charged, creating a symmetric electric field.
The formula to calculate the electric field just above the middle of the sheet is:
E = σ / (2ε₀)
Where σ is the charge density and ε₀ is the permittivity of free space, which is approximately 8.85 x 10⁻¹² C²/(N·m²).
Substituting the given values, we have:
E = (5.40 x 10⁻⁶ C/m²) / (2 x 8.85 x 10⁻¹² C²/(N·m²))
Simplifying, we get:
E = 3.05 x 10⁶ N/C
The direction of the electric field is perpendicular to the sheet and points away from it.
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6. calculate the power of the eye when viewing an object 3.00 m away.
The power of the eye when viewing an object 3.00 m away is 58.82 diopters (D).
What is power of the eye?
The power of the eye refers to its ability to refract light and focus it onto the retina, enabling clear vision at different distances. The power of the eye is measured in diopters (D).
In the case of the human eye, it is generally assumed to have an equivalent focal length of approximately 17 mm, which is equivalent to 0.017 m. This value represents the average refractive power of the eye.
To calculate the power of the eye when viewing an object 3.00 m away, we can use the following formula:
P = 1 / f
P = 1 / 0.017
P = 58.82 D
Therefore, the power of the eye when viewing an object 3.00 m away is approximately 58.82 diopters (D).
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if 104 w/m2 corresponds to 160 db, what is the sound intensity level, in decibels, of ultrasound with intensity 10^5 w/m2, used to pulverize tissue during surgery?
β= ___ dB
The sound intensity level of the ultrasound used to pulverize tissue during surgery is approximately 29.83 dB.
To determine the sound intensity level (β) in decibels (dB) for ultrasound with an intensity of ([tex]10^{5}[/tex] W/[tex]m^{2}[/tex]), we can use the formula for sound intensity level
β = 10 * log10(I/I₀)
Where:
β is the sound intensity level in decibels,
I is the sound intensity of the ultrasound,
I₀ is the reference sound intensity.
In this case, the reference sound intensity corresponds to 104 W/[tex]m^{2}[/tex], which corresponds to 160 dB. Therefore, we have:
β = 10 * log10(I / 104)
Plugging in the given intensity of ultrasound ([tex]10^{5}[/tex] W/[tex]m^{2}[/tex]), we have:
β = 10 * log10([tex]10^{5}[/tex] / 104)
β = 10 * log10(961.54)
β = 10 * 2.983
β = 29.83 dB
Therefore, the sound intensity level of the ultrasound used to pulverize tissue during surgery is approximately 29.83 dB.
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Fill in the blank with either “insulators” or “conductors” to complete the sentence. The conductivity of a material is determined by the number of free electrons. have few or no free electrons.
Answer:
Insulators is the correct answer
Explanation:
A few electrons make up the insulators, which allow for very little electrical current to flow. While the conductors are excellent electrical conductors. Rubber is a poor insulator compared to metal, which is a very good conductor.
What are insulators?The insulator is a material that consists of a few electrons and they permit very little flow of electrical current. While the conductors are very good conductors of electricity. Metal is a very good conductor while rubber is a poor insulator.
Hence the answer is conductors possess free electrons, while insulators have few free electrons
The conductor has thus been used for wiring purposes and can be effectively used made to use energy; The insulator is a poor absorber and thus are unable to transfer energy.
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two charged particles held near each other are released. as they move, the acceleration of each decreases. therefore, the particles have
Explanation:
LIKE charge..... if they were opposite (opposites attract) ...they would accelerate as they grew closer .....like charges REPEL and they get farther apart and decelerate ....
Mario uses a hot plate to heat a beaker of 150mL of water. He used a
thermometer to measure the temperature of the water. The water in the
beaker began to boil when it reached the temperature of 100 degrees
Celsius. If Mario completes the same experiment with 200 mL of water,
what would happen to the boiling point? *
Answer:
the point of the BOILING point DOES NOT CHANGE
T = 100ºC
Explanation:
The boiling point of a substance is when it changes from a liquid to a gaseous state.
This point depends on the forces that join the molecules of the liquid, if we take the macroscopic variables of thermodynamics: Temperature, pressure depends on the two in a phase diagram, but nowhere is it dependent on the amount of matter.
Therefore the point of the BOILING point DOES NOT CHANGE
T = 100ºC
Show that liquid pressure is directly
proportional to height of liquid in a vessel.
Answer:
P=F/A where F is the weight of the water and A is the area on which it is resting. The weight of the water is mg. The mass of the water is dv where d is the density and v is the volume. Finally, the volume of the water in a vessel is equal to the area of the base of the vessel times the height of the vessel. (v=Ah)
Plugging everything in we get:
P = dAhg/A
So
P=dhg
So we have shown that liquid pressure is directly proportional to height of liquid in a vessel.
f. write two reasons why it is better to obtain the moment of inertia through a linear fit than by solving for i in the equation and plugging in the value of ωmax and θ0 for one or two points.
Two reasons why it is better to obtain the moment of inertia through a linear fit rather than by solving for I using specific points (ωmax and θ0) are:
Increased Precision: A linear fit allows for the consideration of a larger set of data points, which can provide a more accurate determination of the moment of inertia. By analyzing the relationship between θ and ω over a range of values, the linear fit captures the overall trend and minimizes the potential errors associated with individual data points. This leads to a more precise estimation of the moment of inertia compared to relying on only one or two specific points. Account for Nonlinearities: In some cases, the relationship between θ and ω may not follow a simple linear pattern. If nonlinearity exists, using a linear fit provides a more flexible approach to capture the overall trend. By fitting a line to the data, even if the relationship is not strictly linear, we can still obtain a reasonable approximation of the moment of inertia by considering the best-fit line that represents the general behavior of the system. This method accounts for potential nonlinearities and provides a more reliable estimate of the moment of inertia compared to a limited number of specific points.
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A 5000kg elephant steps into a large spring and compresses it from 1m long to 50cm long what is the spring constant of the spring
Answer: 98 kN/m
Explanation:
Given
Mass of elephant [tex]m=5000\ kg[/tex]
Spring compresses from [tex]1\ m\ (100\ cm) \text{to}\ 50\ cm[/tex]
i.e. change in length is [tex]100-50=50\ cm[/tex]
spring force is given by [tex]kx[/tex]
where k=spring constant
x=change in length
The weight of elephant must be equal to the spring force
[tex]\Rightarrow W=kx\\\Rightarrow 5000\times 9.8=k\times 0.5\\\Rightarrow k=98,000\ N/m\ or\ 98\ kN/m[/tex]
) A spring with a spring constant of 2.1N/m is stretched 0.2m. What force is produced by
the spring?
Answer:
Force = 0.42 Newton
Explanation:
Given the following data;
Spring constant, k = 2.1
Extension, e = 0.2m
To find the force, we would use the formula below;
Force = spring constant * extension
Force = 2.1 * 0.2
Force = 0.42 Newton
A totally reflecting disk has radius 8. 00 um, thickness 2. 00 um, and average density 7. 00x102 kg/m². A laser has an average power output Pav spread uniformly over a cylindrical beam of radius 2. 00 mm. When the laser beam shines upward on the disk in a direction perpendicular to its flat surface, the radiation pressure produces a force equal to the weight of the disk. Part A What value of Pay is required? Express your answer with the appropriate units. ΜΑ ?
Answer:
2.69 × 10^(-10) N
Explanation:
To calculate the required value of power, we need to consider the force exerted by the radiation pressure and equate it to the weight of the disk.
The force exerted by the radiation pressure is given by:
F = (2RΔt)P/c
where:
F is the force,
R is the radius of the disk,
Δt is the thickness of the disk,
P is the power of the laser,
c is the speed of light.
We are given:
R = 8.00 μm = 8.00 × 10^(-6) m (radius of the disk)
Δt = 2.00 μm = 2.00 × 10^(-6) m (thickness of the disk)
ρ = 7.00 × 10^2 kg/m² (average density of the disk)
The weight of the disk is given by:
W = mg
where:
m is the mass of the disk,
g is the acceleration due to gravity.
The mass of the disk can be calculated using its average density and volume:
m = ρV
The volume of the disk is given by:
V = πR²Δt
Substituting the expressions for mass and volume into the equation for weight, we have:
W = ρVg = ρ(πR²Δt)g
Setting the force equal to the weight, we have:
F = W
(2RΔt)P/c = ρ(πR²Δt)g
Simplifying the equation:
2RP/c = ρπR²g
Now we can solve for the power P:
P = (ρπRg)/(2c)
Substituting the given values:
P = (7.00 × 10^2 kg/m²)(π)(8.00 × 10^(-6) m)(9.8 m/s²)/(2(3.00 × 10^8 m/s))
Calculating this expression:
P ≈ 2.69 × 10^(-10) kg⋅m/s² = 2.69 × 10^(-10) N
So, the required power Pay is approximately 2.69 × 10^(-10) N (newtons).
Define Eukaryote cells and Prokaryote cells
Answer:
Ekaryote cells
an organism consisting of a cell or cells in which the genetic material is DNA in the form of chromosomes contained within a distinct nucleus. Eukaryotes include all living organisms other than the eubacteria and archaea.
prokaryote cells
microscopic single-celled organism which has neither a distinct nucleus with a membrane nor other specialized organelles, including the bacteria and cyanobacteria
Why do baseball pitchers throw the ball at an angle that is slightly above the horizontal if they want the ball to reach at approximately the same height as it was thrown when it gets to the batter?
Answer:
The angle above the horizontal at which the pitcher throws the ball determines the distance the ball travels before returning to the height at which it was thrown
Explanation:
The baseball is thrown as a projectile and the range, 'R', of the baseball which is the distance the baseball travels before the height above the ground returns to the initial height is given given as follows;
[tex]R = \dfrac{u^2 \cdot sin(2\cdot \theta )}{g}[/tex]
Where;
R = The range of the baseball = The horizontal distance away from the pitcher the ball reaches
u = The initial velocity with which the baseball was thrown
θ = The angle above horizontal a baseball pitcher throws the ball
g = The acceleration due to gravity ≈ 9.81 m/s²
From the the equation, when θ = 0, sin(θ) = sin(0) = 0 and the ball does not cover any horizontal distance before going lower than the height at which it was thrown, therefore, for the ball to travel further, the angle of launch, θ has to be larger than 0.
The point of this question is to compare rest energy and kinetic energy at low speeds. A baseball is moving at a speed of 34 m/s. Its mass is 145 g (0.145 kg)
a) What is its rest energy?
b) What is its kinetic energy?
c) Which of the following statements is true:
i) The kinetic energy is much bigger than the rest energy.
ii) The kinetic energy is approximately equal to the rest energy.
iii) The kinetic energy is much smaller than the rest energy.
a) The rest energy (Eₙ) is 1.305 x 10¹⁶ J.
b) The kinetic energy (Eₖ) is 83.71 J.
c) The statement that is true is " The kinetic energy is much smaller than the rest energy".
To compare the rest energy and kinetic energy of the baseball, we can use the equations:
a) The rest energy (Eₙ) can be calculated using Einstein's mass-energy equivalence formula:
Eₙ = m × c²
Where:
m is the mass of the baseball
c is the speed of light in a vacuum, which is approximately 3.00 x 10⁸ m/s
Given:
m = 0.145 kg
Using the formula:
Eₙ = 0.145 kg × (3.00 x 10⁸ m/s)²
Eₙ = 0.145 kg × 9.00 x 10¹⁶ m²/s²
Eₙ = 1.305 x 10¹⁶ J
b) The kinetic energy (Eₖ) can be calculated using the formula:
Eₖ = (1/2) × m × v²
Where:
m is the mass of the baseball
v is the velocity of the baseball
Given:
m = 0.145 kg
v = 34 m/s
Using the formula:
Eₖ = (1/2) × 0.145 kg × (34 m/s)²
Eₖ = 0.0725 kg × (1156 m²/s²)
Eₖ = 83.71 J
c) Comparing the rest energy and kinetic energy:
i) The kinetic energy is much bigger than the rest energy.
ii) The kinetic energy is approximately equal to the rest energy.
iii) The kinetic energy is much smaller than the rest energy.
In this case, the rest energy is approximately 1.305 x 10¹⁶ J, while the kinetic energy is 83.71 J. Therefore, statement iii) is true: the kinetic energy is much smaller than the rest energy.
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The voltage v (in volts) induced in a tape head is given by v = t^2 e^3t, where t is the time (in seconds). Find the average value of v over the interval from t = 0 to t = 3. Round to the nearest volt. A) 16 volts B) 1100 volts C) 6502 volts D) 71, 327 volts
The average value of the voltage induced in a tape head over the interval from t = 0 to t = 3 can be found by evaluating the definite integral of the given function. Rounded to the nearest volt, the answer is C) 6502 volts.
To find the average value of the voltage v over the interval from t = 0 to t = 3, we need to evaluate the definite integral of the function [tex]v = t^2 e^3^t[/tex] with respect to t over this interval. The average value is given by the formula:
Average value = (1 / (b - a)) * ∫(a to b) v dt,
where a and b represent the starting and ending points of the interval, respectively.
Substituting the given values a = 0 and b = 3 into the formula, we have:
Average value = (1 / (3 - 0)) * ∫(0 to 3) [tex]t^2 e^3^t[/tex]dt.
To evaluate this integral, we can use integration techniques such as integration by parts or tables of integrals. Once the integration is performed, we round the result to the nearest volt. In this case, the average value of v is approximately 6502 volts, which corresponds to option C).
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If a ray of light makes a 45° from the normal incident from air (n=1) into glass (n=1.7) what is the angle of the refracted ray from the normal? Select one: a. 1.2 b. 17° c. 24.6° d. 45°
The angle of the refracted ray from the normal is 24.6°.
The correct answer is option c. 24.6°.
When a ray of light travels from one medium to another, it bends as its speed changes due to the change in medium. This bending is called refraction. The angle of incidence (θ1) is the angle between the incident ray and the normal, while the angle of refraction (θ2) is the angle between the refracted ray and the normal.Here,
the angle of incidence is 45°,
the refractive index of air (n1) is 1 and
the refractive index of glass (n2) is 1.7.
Using Snell's Law,
n1 sin(θ1) = n2 sin(θ2)
Substituting the values,
1 × sin(45°) = 1.7 × sin(θ2)
sin(θ2) = (1 × sin(45°)) / 1.7
sin(θ2) = 0.4482
θ2 = sin^-1(0.4482)
θ2 = 24.6°
Therefore, the angle of the refracted ray from the normal is 24.6°.
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How much work w must be done on a particle with a mass of m to a\ccelerate it from rest to a speed of 0.902 c ? express your answer as a multiple of mc2 to three significant figures.
We can utilize Einstein's mass-energy equivalence equation, E = mc², where E represents the energy. The work done on the particle is equal to the change in energy.
When the particle is at rest, its energy is solely its rest energy, which is given by E = mc². As the particle is accelerated to a speed of 0.902 c, its total energy increases. The change in energy (ΔE) is the difference between the final energy and the initial rest energy.
The final energy of the particle when it reaches a speed of 0.902 c is given by E = γmc², where γ is the Lorentz factor. The Lorentz factor is defined as γ = 1/√(1 - (v/c)²), where v is the velocity of the particle.
By substituting the given values into the Lorentz factor equation, we can calculate the Lorentz factor for the particle. With the Lorentz factor known, we can determine the final energy of the particle.
The work done on the particle is equal to the change in energy, so the work can be calculated as ΔE = (γ - 1)mc². By substituting the values into the equation and expressing the answer as a multiple of mc², we can determine the work required to accelerate the particle to the given speed.
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ANSWER ASAPPP
What type of mountain would form from vertical movements along fault lines?
A) volcano
B) dome
C) fault-block
D) folded
C) fault-block mountain
at a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.
A)73 °C
B)379 °C
C)32 °C
D)14 °C
E)106 °C
The ideal gas law can be used to solve the problem. The relationship between pressure, temperature, volume, and the number of moles of gas is given by the ideal gas law. PV = nRT is the formula for the ideal gas law, Where:P = pressure in atm, V = volume in Litersn = a number of moles, R = gas constant, T = temperature in Kelvin.
The temperature at which 0.444 mol of CO gas occupies 11.8 L at 889 torr can be calculated as follows ;
First, we must convert the pressure from torr to atm: 889 torr × 1 atm/760 torr = 1.17 atm.
Using the ideal gas law formula: PV = nRT.
We can solve for T.T = PV/nR where, P = 1.17 atmV = 11.8 Ln = 0.444 molR = 0.08206 L•atm/mol•K.
Substitute the given values,T = (1.17 atm × 11.8 L)/(0.444 mol × 0.08206 L•atm/mol•K)T = 379 K.
Convert Kelvin to Celsius by subtracting 273.15 from the value,379 K - 273.15 = 105.85°C ≈ 106°C.
Therefore, the temperature at which 0.444 mol of CO gas occupies 11.8 L at 889 torr is approximately 106°C. Hence, the correct option is E)106 °C.
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A wedge with a mechanical advantage of 0.78 is used to raise a house corner from its foundation. If the output force is 7500 N, what is the input force?
Therefore, the input force is 9615.38N
Input force calculation.
T0 determine the input force needed 0.78 to raise a house corner from its foundation we need to use the formula for mechanical advantage.
Mechanical advantage = output force/ input force
Given the output force to be 7500N and the mechanical advantage 0.78.
We can rearrange the the formula
IF = 7500/0.78
IF = 9615.38N
Therefore, the input force is 9615.38N
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When copper combines with oxygen to form copper(II) oxide, the charge of the copper ion is
A conveyer belt carries a load of mass 180kg n lift it up in 1
Answer:
Cool.
Explanation:
What's the question..? :|
Question 16 (1 point)
Light travels through a material at a speed of 4.3 x 108 m/s. What is the index of
refraction for the material?
Your Answer:
Answer
units
Answer:
η = 0.7
Explanation:
The refractive index of the material can be calculated by using the following formula:
[tex]\eta = \frac{c}{v}\\[/tex]
where,
η = refractive index of the material = ?
c = speedof light in vaccuum = 3 x 10⁸ m/s
v = speed of light in this material = 4.3 x 10⁸ m/s
Therefore,
[tex]\eta = \frac{3\ x\ 10^8\ m/s}{4.3\ x\ 10^8\ m/s}[/tex]
η = 0.7
If you're driving one and a half miles per minute, slow down by 15 miles per hour, and then reduce your speed by one third, how fast are you going now?
a.90 miles per hour
b.60 miles per hour
c.50 miles per hour
d.75 miles per hour
e.45 miles per hour
After slowing down by 15 miles per hour and reducing the speed by one third, you are now going at 60 miles per hour.
What is the current speed after slowing down?If you are initially driving at a rate of one and a half miles per minute, it means you are traveling at a speed of 90 miles per hour (since there are 60 minutes in an hour).
In the second step, you slow down by 15 miles per hour. This reduces your speed to 75 miles per hour.
Finally, you reduce your speed by one third, which means you need to subtract one third of 75 from 75. One third of 75 is 25, so subtracting 25 from 75 gives you a speed of 50 miles per hour.
Therefore, after slowing down by 15 miles per hour and then reducing your speed by one third, you are now traveling at a speed of 50 miles per hour.
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A 3.0-kg ball with an initial velocity of (4i +3j) m/s collides with a wall and rebounds with avelocity of (-4i + 3j) m/s.What is the impulse exerted on the ball by the wall?
Question 2 answers
a. +24i N s
b. -24i N s
c. +18j N s
d. -18j N s
e. +8.0i N s
The impulse exerted on the ball by the wall is -24i N s, the answer is option b.
The impulse exerted on the ball by the wall can be determined using the impulse-momentum theorem.
Impulse is equal to the change in momentum.
That is,
J = Δp
Since momentum is a vector,
p = mv
Substituting the values,
p(initial) = mv(initial) = 3.0 kg × (4i + 3j) m/s
p(initial) = 12i + 9j N s
p(final) = mv(final) = 3.0 kg × (-4i + 3j) m/s
p(final) = -12i + 9j N s
Change in momentum,
(Δp) = p(final) - p(initial)
= (-12i + 9j) N s - (12i + 9j) N s
= (-12 - 12)i + (9 - 9)j N s
= -24i N s
Thus, the impulse exerted on the ball by the wall is -24i N s, which means that the answer is option b.
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