Answer:
It becomes wider
Explanation:
Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.
Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?
Answer:
The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.
A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
The expected year is 2017.
Explanation:
Total years that the millionaire to live = 15 years
Travel away from the earth at = 0.8 c
This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:
[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]
Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017
A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm
Answer:
Explanation:
The double slit interference phonemene is described for the case of constructive interference
d sin θ= m λ (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
The double slit interference phonemene is described for the case of constructive interference
d sin θ = m lam (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in equation 1
d y / L = m λ
λ = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 mm = 0.44 10⁻³ m
y = 5.5 mm = 5.5 10⁻³ m
L = 4.2m
m = 1
let's calculate
λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
λ = 5.76190 10-7 m
let's reduce to num
lam = 5.56190 10-7 m (109 nm / 1m)
lam = 556,190 nmtea
we substitute
without tea = y / L
we substitute in equation 1
d y / L = m lam
lam = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 me = 0.44 10-3 m
y = 5.5 mm = 5.5 10-3
L = 4.2m
m = 1
let's calculate
lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
lam = 5.76190 10⁻⁷ m
let's reduce to num
lam = 5.56190 10⁻⁷ m (109 nm / 1m)
lam = 556,190 nm
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]
Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.
Answer:
Zack should direct his throw outward and toward the back of the car.
Explanation:
As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.
The solution is throw 3.
I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Which statement best applies Newton’s laws of motion?The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.
When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.
The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.
Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
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A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
A computer has a mass of 3 kg. What is the weight of the computer?
A. 288 N.
B. 77.2 N
C. 3N
D. 29.4 N
Answer:
29.4 NOption D is the correct option.
Explanation:
Given,
Mass ( m ) = 3 kg
Acceleration due to gravity ( g ) = 9.8 m/s²
Weight ( w ) = ?
Now, let's find the weight :
[tex]w \: = \: m \times g[/tex]
plug the values
[tex] = 3 \times 9.8[/tex]
Multiply the numbers
[tex] = 29.4 \: [/tex] Newton
Hope this helps!!
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The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident upon a rectangular area, 1.5 m by 2.0 m, at right angles. How much total electromagnetic energy falls on the area during 1.0 minute
Answer:
The total energy is [tex]T = 169.02 \ J[/tex]
Explanation:
From the question we are told that
The Poynting vector (energy flux ) is [tex]k = 0.939 \ W/m^2[/tex]
The length of the rectangle is [tex]l = 1.5 \ m[/tex]
The width of the rectangle is [tex]w = 2.0 \ m[/tex]
The time taken is [tex]t = 1 \ minute = 60 \ s[/tex]
The total electromagnetic energy falls on the area is mathematically represented as
[tex]T = k * A * t[/tex]
Where A is the area of the rectangle which is mathematically represented as
[tex]A= l * w[/tex]
substituting values
[tex]A= 2 * 1.5[/tex]
[tex]A= 3 \ m^2[/tex]
substituting values
[tex]T = 0.939 * 3 * 60[/tex]
[tex]T = 169.02 \ J[/tex]
Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations.
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)
Answer:
A) 1.76U×10⁻³N
B) 2.716×10⁻³N
C) 264.5⁰
Explanation:
See detailed workings for (a), (b), (c) attached.
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness
Answer:
A. NearsightednessExplanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given by
E=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Answer:
Explanation:
E= −L ΔI / Δt.
L = E Δt / ΔI
Hence the unit of inductance may be V s A⁻¹
or volt s per ampere .
In the given case
change in current ΔI = - 2.5 A
change in time = .015 s
L = .56 H
E = − L ΔI / Δt.
= .56 x 2.5 / .015
= 93.33 V .
Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
6.371 * 10^6 m and the gravitational acceleration at the surface of the Earth is 9.81
m/s^2 what is the gravitational acceleration at the surface of an alien planet with
2.4 times the mass of the Earth and 1.9 times the radius of the Earth?
Although you do not necessarily need it the universal gravitational constant is G =
6.674 * 10^(-11) N*m^2/kg^2
9
Answer:
gₓ = 6.52 m/s²
Explanation:
The value of acceleration due to gravity on the surface of earth is given as:
g = GM/R² -------------------- equation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
M = Mass of Earth
R = Radius of Earth
Now, for the alien planet:
gₓ = GMₓ/Rₓ²
where,
gₓ = acceleration due to gravity at the surface of alien planet
Mₓ = Mass of Alien Planet = 2.4 M
Rₓ = Radius of Alien Planet = 1.9 R
Therefore,
gₓ = G(2.4 M)/(1.9 R)²
gₓ = 0.66 GM/R²
using equation 1
gₓ = 0.66 g
gₓ = (0.66)(9.81 m/s²)
gₓ = 6.52 m/s²
The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?
Answer:
1.03A
Explanation:
For computing the magnitude of the current in the circuit we need to do the following calculations
LCR circuit impedance
[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]
= 148.7Ω
Now the phase angle is
[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]
Now the rms current flowing in the circuit is
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]
= 0.98 A
The current flowing in the circuit is
[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]
= 1.39 A
And, finally, the current across the generator is
[tex]I'= I cos \phi[/tex]
[tex]= (1.39) cos 42.3^{\circ}[/tex]
= 1.03A
Hence, the magnitude of the circuit current is 1.03A
dandre expands 120w of power in moving a couch 15 meters in 5 seconds how much force does he exert ?
Answer:
The answer is 40 N for APX
Explanation:
A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before pC
after pC
(b) Determine the capacitance and potential difference after immersion.
Cf = F
ΔVf = V
(c) Determine the change in energy of the capacitor.
[ ] nJ
Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the food, how should the person aim the arrow containing the food?
Answer:
The options are
A. He should aim it at the monkey
B. He should aim it below the monkey
C. He should aim it above the monkey
D. None of the above
The answer is A. He should aim it at the monkey
This is because the monkey has a large surface area and a bigger body mass. This will make aiming the food at the monkey feasible in it getting it as it could use other parts of the body to get the food aimed at it. The monkey won’t reach the food when falling if it is aimed above it. It also won’t get to the it when it is shot at below it.
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same.
A)
The new sphere has a density of ρ = ρ0 and a mass of m < m0.
B)
The new sphere has a density of ρ = ρ0 and a radius of r > r0.
C)
The new sphere has a density of ρ < ρ0 and a mass of m = m0.
The options are r, f, and s. Rises, Falls, Stays the same.
Answer:
(a) f
(b) r
(c) s
Explanation:
There are two forces on the sphere: weight and buoyancy.
Sum of forces in the y direction:
∑F = ma
B − mg = 0
B = mg
Buoyancy is equal to the weight of the displaced fluid, or ρVg, where ρ is the density of the fluid and V is the displaced volume.
ρVg = mg
ρV = m
V = m/ρ
(a) The mass decreases, so the displaced volume decreases.
(b) The sphere's density is constant and its radius increases, which means its mass increases, so the displaced volume increases.
(c) The mass stays the same, so the displaced volume is the same.
have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons
Answer:
B
Explanation:
Protons have a positive electrical charge of +1,
Electrons have a negative charge of -1,
Neutrons have a neutral charge of about 0.
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .
Answer:
The weight of the rod is 32.87 N
Explanation:
Density of the rod = 7800 kg/m
length of the rod = 81.2 cm = 0.812 m
diameter of rod = 2.60 cm = 0.026 m
acceleration due to gravity = 9.80 m/s^2
The rod can be assumed to be a cylinder.
The volume of the rod can be calculated as that of a cylinder, and can be gotten as
V = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter of the rod
l is the length of the rod
V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3
We know that the mass of a substance is the density times the volume i.e
mass m = ρV
where ρ is the density of the rod
V is the volume of the rod
m = 4.3 x 10^-4 x 7800 = 3.354 kg
The weight of a substance is the mass times the acceleration due to gravity
W = mg
where g is the acceleration due to gravity g = 9.80 m/s^2
The weight of the rod W = 3.354 x 9.80 = 32.87 N
A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
[tex]F = \frac{GMm}{r^2}[/tex]
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]
Gravitational force between the second object and the 42 kg object
[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
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When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the copper is cooled with liquid nitrogen, the resistivity of the copper drops dramatically. How will this affect the speed at which the magnet falls toward the copper
Answer:
The speed at which the magnet falls through the copper block will be reduced dramatically.
Explanation:
Eddy's current are loops of electrical current induced within conductors by a changing magnetic field in the conductor. Eddy's current is proportional to the the magnetic field strength, the rate of change of flux, the area of the loop, and is inversely proportional to the resistivity of the material. Eddy currents flows perpendicularly to the magnetic field, and in closed loops within conductors.
Reducing the resistivity of the copper will increase the Eddy current on the copper, which will in turn increase the opposition to the action producing the flux change (the falling magnet through the copper block). The result is that the speed at which the magnet falls through the copper block will be reduced dramatically.
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
gravity can be described as..?
A. an magnetic force found in nature
B.the force that moves electrical charges
C.the force that repels object with like chargers
D.the force of attraction between two objects
Answer:
D
Explanation:
Gravity is the force of attraction between two objects.
Each object creates a gravitational field in wich every other object is affected by it.
The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
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A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
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An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a
Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a
Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.
Answer:
E = α/2∈₀ [ 1 - a²/r² ]
Ф = α/2∈₀
Explanation:
Using Gauss Law:
ρ(r) = a/r, dA
= 4 π r²d r
Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA
Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA
= 4πα [tex]\int\limits^r_a[/tex] r'dr'
Ф[tex]_{encl}[/tex] = 4 π α 1/2(r²-a²)
E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀
= [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)
= α (r² - a²) / 2 ∈₀ (r²)
= α/2∈₀ [ r²/r² - a²/r² ]
E = α/2∈₀ [ 1 - a²/r² ]
Electric field of the point charge:
E[tex]_{q}[/tex] = q / 4π∈₀r²
[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²
For [tex]E_{total}[/tex] to be constant:
- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²
-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀
= α - αa² + αa² / 2 ∈₀
= α /2 ∈₀
Hence:
Ф = α/2∈₀
As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the index of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21
Explanation:
It is given that,
Angle of incidence from air to another medium, i = 26°
The angle of reflection, r = 32°
We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :
[tex]n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82[/tex]
So, the index of refraction is 0.82. Hence, the correct option is C.