Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = chargeA very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on both objects.
Answer:
In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Explanation:
In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.
Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
Answer:
x=22.57 m
Explanation:
Given that
35 m in W of S
angle = 40 degrees
25 m in east
From the diagram
The angle
[tex]\theta=90-40=50^o[/tex]
From the triangle OAB
[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]
[tex]1340.57=35^2+25^2-x^2[/tex]
x=22.57 m
Therefore the answer of the above problem will be 22.57 m
can I get help please?
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacitor?Now a conductor is inserted into the capacitor. The thickness of the conductor is 1/3 the distance between the plates of the capacitor and is centered inbetween the plates of the capacitor.b) What is the charge on the plates of the capacitor?c) What is the capacitance of the capacitor with the conductor in place?d) What is the energy stored in the capacitor with the conductor in place?
Answer:
a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]
b) Q = 45 µC
c) C' = 1.5 μF
d) [tex]E = 6.75 *10^{-4} J[/tex]
Explanation:
Capacitance, C = 1 µF
Charge on the plates, Q = 45 µC
a) Energy stored in the capacitor is given by the formula:
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]
b) The charge on the plates of the capacitor will not change
It will still remains, Q = 45 µC
c) Electric field is non zero over (1-1/3) = 2/3 of d
From the relation V = Ed,
The voltage has changed by a factor of 2/3
Since the capacitance is given as C = Q/V
The new capacitance with the conductor in place, C' = (3/2) C
C' = (3/2) * 1μF
C' = 1.5 μF
d) Energy stored in the capacitor with the conductor in place
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]
An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2. The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the direction of the cliff.
Required:
a. Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight.
b. The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the cliff the coyote lands.
c. Determine the components of the coyote’s impact velocity
Answer:
a) v_correcaminos = 22.95 m / s , b) x = 512.4 m ,
c) v = (45.83 i ^ -109.56 j ^) m / s
Explanation:
We can solve this exercise using the kinematics equations
a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff
v² = v₀² + 2 a x
they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²
v = √ (2 15 70)
v = 45.83 m / s
with this value calculate the time it takes to arrive
v = v₀ + a t
t = v / a
t = 45.83 / 15
t = 3.05 s
having the distance to the cliff and the time, we can find the constant speed of the roadrunner
v_ roadrunner = x / t
v_correcaminos = 70 / 3,05
v_correcaminos = 22.95 m / s
b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.
Let's start by looking for the time to reach the cliff floor
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
in this case y = 0 and the height of the cliff is y₀ = 100 m
0 = 100 + 45.83 t - ½ 9.8 t²
t² - 9,353 t - 20,408 = 0
we solve the quadratic equation
t = [9,353 ±√ (9,353² + 4 20,408)] / 2
t = [9,353 ± 13] / 2
t₁ = 11.18 s
t₂ = -1.8 s
Since time must be a positive quantity, the answer is t = 11.18 s
we calculate the horizontal distance traveled
x = v₀ₓ t
x = 45.83 11.18
x = 512.4 m
c) speed when it hits the ground
vₓ = v₀ₓ = 45.83 m / s
we look for vertical speed
v_{y} = [tex]v_{oy}[/tex] - gt
v_{y} = 0 - 9.8 11.18
v_{y} = - 109.56 m / s
v = (45.83 i ^ -109.56 j ^) m / s
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.
Question:
A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)
Answer:
3.22 x 10⁻⁴ m/s
Explanation:
The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;
v = [tex]\frac{I}{nqA}[/tex]
Where;
n = number of free electrons per cubic meter
q = electron charge
A = cross-sectional area of the wire
First let's calculate the number of free electrons per cubic meter (n)
Known constants:
density of copper, ρ = 8.95 x 10³kg/m³
molar mass of copper, M = 63.5 x 10⁻³kg/mol
Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol
But;
The number of copper atoms, N, per cubic meter is given by;
N = (Nₐ x ρ / M) -------------(ii)
Substitute the values of Nₐ, ρ and M into equation (ii) as follows;
N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³
N = 8.49 x 10²⁸ atom/m³
Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;
n = 8.49 x 10²⁸ electrons/m³
Now let's calculate the drift electron
Known values from question:
A = 5.261 mm² = 5.261 x 10⁻⁶m²
I = 23A
q = 1.6 x 10⁻¹⁹C
Substitute these values into equation (i) as follows;
v = [tex]\frac{I}{nqA}[/tex]
v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]
v = 3.22 x 10⁻⁴ m/s
Therefore, the drift electron is 3.22 x 10⁻⁴ m/s
Observe the process by which the grey and the red spheres are charged using the electrophorus. After each sphere is first charged, what are their charges
Answer:
The gray spheres is negatively charged while the red is positively charged
Explanation:
This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged
Answer:
The gray sphere has a positive charge and the red sphere has a positive charge.
At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!
An ice skater is in a fast spin with her arms held tightly to her body. When she extends her arms, which of the following statements in NOT true?
A. Het total angular momentum has decreased
B. She increases her moment of inertia
C. She decreases her angular speed
D. Her moment of inertia changes
Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Answer:
[tex]v_{2}=3.5 m/s[/tex]
Explanation:
Using the conservation of energy we have:
[tex]\frac{1}{2}mv^{2}=mgh[/tex]
Let's solve it for v:
[tex]v=\sqrt{2gh}[/tex]
So the speed at the lowest point is [tex]v=7 m/s[/tex]
Now, using the conservation of momentum we have:
[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
[tex]v_{2}=\frac{1*7}{2}[/tex]
Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]
I hope it helps you!
A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 29.0 m. After what time interval does it strike the ground?
Answer: after 1.75 seconds
Explanation:
The only force acting on the ball is the gravitational force, so the acceleration will be:
a = -9.8 m/s^2
the velocity can be obtained by integrating over time:
v = -9.8m/s^2*t + v0
where v0 is the initial velocity; v0 = -7.95 m/s.
v = -9.8m/s^2*t - 7.95 m/s.
For the position we integrate again:
p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0
where p0 is the initial position: p0 = 29m
p = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Now we want to find the time such that the position is equal to zero:
0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Then we solve the Bhaskara's equation:
[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]
Then the solutions are:
t = (7.95 + 25.1)/(-9.8) = -3.37s
t = (7.95 - 25.1)/(-9.8) = 1.75s
We need the positive time, then the correct answer is 1.75s
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 N.m. When the dipole reaches equilibrium its potential energy is equal to -0.2 J. What was the initial angle between the direction of the dipole moment and the direction of the electric field?
Answer:
θ = 180
Explanation:
When an electric dipole is placed in an electric field, there is a torque due to the electric force
τ = p x E
by rotating the dipole there is a change in potential energy
ΔU = ∫ τ dθ
ΔU = p E (cos θ₂ - cos θ₁)
when the dipole starts from an angle to the equilibrium position for θ = 0
ΔU = pE (cos θ - cos 0)
cos θ = 1 + DU / pE)
let's apply this expression to our case, the change in potential energy is ΔU = -0.2J
let's calculate
cos θ = 1 -0.2 / 0.1
cos θ = -1
θ = 180
A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about point A, that is, the point of contact between the front tire of the forklift and the ground
Answer:
The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]
Explanation:
mass of the crate = [tex]M_{C}[/tex]
speed of forklift = [tex]V_{1}[/tex]
The distance between the center of the mass and the point A = d
Recall that the angular moment is the moment of the momentum.
[tex]L = P*d[/tex] ..... equ 1
where L is the angular momentum,
P is the momentum of the system,
d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image
Also, we know that the momentum P is the product of mass and velocity
P = mv ....equ 2
in this case, the mass = [tex]M_{C}[/tex]
the velocity = [tex]V_{1}[/tex]
therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]
we substitute equation 2 into equation 1 to give
[tex]L = M_{C} V_{1} d[/tex]
A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed
Answer:
The new angular speed is [tex]w = 6 \ rev/s[/tex]
Explanation:
From the question we are told that
The angular velocity of the spin is [tex]w_o = 3 \ rev/s[/tex]
The original moment of inertia is [tex]I_o[/tex]
The new moment of inertia is [tex]I =\frac{I_o}{2}[/tex]
Generally angular momentum is mathematically represented as
[tex]L = I * w[/tex]
Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so
[tex]I * w = constant[/tex]
=> [tex]I_o * w _o = I * w[/tex]
where w is the new angular speed
So
[tex]I_o * 3 = \frac{I_o}{2} * w[/tex]
=> [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]
=> [tex]w = 6 \ rev/s[/tex]
If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?
Complete Question:
A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.
(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its
speed is 3.7 m/s, what angle does the cord make with the vertical?
(Check attached image for the diagram.)
Answer:
(a) The ball’s speed, v = 2.06 m/s
(b) The angle the cord makes with the vertical is 50.40⁰
Explanation:
If the ball is revolved in a horizontal plane, it will form a circular trajectory,
the radius of the circle, R = Lsinθ
where;
L is length of the string
The force acting on the ball is given as;
F = mgtanθ
This above is also equal to centripetal force;
[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]
(b) when the speed is 3.7 m/s
[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]
[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]
Therefore, the angle the cord makes with the vertical is 50.40⁰
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.
Answer:
Velocity.
Explanation:
Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.
As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:
Horizontal range: As per expression:
R= ([tex]u^{2}[/tex]*sin2θ)/g
the range depending on the square of the initial velocity.
Maximum height: As per expression:
H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g
the maximum distance also depends upon square of the initial velocity.
If, the limits of the visible spectrum are approximately 3000 A.U. and 5000 A.U. respectively. Determine the angular breadth of the first order visible spectrum produced by a plane diffraction grating having 12000 lines per inch when light is incident normally on the grating.
Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.
Answer and Explanation:
There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e. [tex]B \alpha \frac{1}{r^2}[/tex]
The magnetic field is a volume of vectors
And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit
Therefore for a closed path, we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.
A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting
Answer:
Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m
Explanation:
The maximum electric field in the beam = 2E
cross-sectional area of beam = A
The intensity of an electromagnetic wave with electric field is
I = cε[tex]E_{0} ^{2}[/tex]/2
for [tex]E_{0}[/tex] = 2E
I = 2cε[tex]E^{2}[/tex] ....equ 1
where
I is the intensity
c is the speed of light
ε is the permeability of free space
[tex]E_{0}[/tex] is electric field
Radiation pressure of an electromagnetic wave on an absorbing surface is given as
P = I/c
substituting for I from above equ 1. we have
P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex] ....equ 2
Also, pressure P = F/A
therefore,
F = PA ....equ 3
where
F is the force
P is pressure
A is cross-sectional area
substitute equ 2 into equ 3, we have
F = 2ε[tex]E^{2}[/tex]A
force on a body = mass x acceleration.
that is
F = ma
therefore,
a = F/m
acceleration of the asteroid will then be
a = 2ε[tex]E^{2}[/tex]A/m
where m is the mass of the asteroid.
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12