Answer:
The voltage across the capacitor will remain constant
The capacitance of the capacitor will increase
The electric field between the plates will remain constant
The charge on the plates will increase
The energy stored in the capacitor will increase
Explanation:
First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.
The charge on the capacitor is equal to
Q = CV
Since the voltage is constant, and the charge increases, the capacitance will also increase.
The energy in a capacitor is given as
E = [tex]\frac{1}{2}CV^{2}[/tex]
since the capacitance has increased, the energy stored will also increase.
Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, and thus the magnetic field it produces. The second coil is connected only to an ammeter. The ammeter will indicate that a current is flowing in the second coil:______.
a. only if the second coil is connected to the power supply by rewiring it to be in series with the first coil.
b. only when the current in the first coil changes.
c. whenever a current flows in the first coil.
d. only when a steady current flows in the first coil.
Answer:
B. only when the current in the first coil changes
Explanation:
This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced
An MRI machine needs to detect signals that oscillate at very high frequencies. It does so with an LC circuit containing a 15mH coil. To what value should the capacitance be set to detect a 450 MHz signal?
Answer:
The capacitance is [tex]C = 3.2 9 *10^{-16} \ F[/tex]
Explanation:
From the question we are told that
The induction of the LC circuit is [tex]L = 15 mH = 15 *10^{-3} \ H[/tex]
The frequency is [tex]w = 450 \ MHz = 450 *10^{6} \ Hz[/tex]
The natural frequency is mathematically represented as
[tex]w = \frac{1}{\sqrt{LC} }[/tex]
Where C is the capacitance So
=> [tex]C = \frac{1}{L * w^2}[/tex]
substituting values
[tex]C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}[/tex]
[tex]C = 3.2 9 *10^{-16} \ F[/tex]
The value for which the capacitance should be set to detect a 450 MHz signal is [tex]8.34 \times 10^{-24} \;F[/tex]
Given the following data:
Inductance = 15 mH = [tex]15 \times 10^{-3}\;H[/tex]Frequency = 450 MHz = [tex]450 \times 10^6 \;Hz[/tex]To determine the value for which the capacitance should be set to detect a 450 MHz signal:
Mathematically, natural frequency is given by the formula:
[tex]f_o = \frac{1}{2\pi \sqrt{LC} }[/tex]
Where:
L is the inductance.C is the capacitance.Making C the subject of formula, we have:
[tex]C = \frac{1}{(2\pi f_o)^2L} \\\\C = \frac{1}{(2\;\times \;3.142 \times \;450 \;\times\; 10^9)^2 \; \times \;15 \times 10^{-3}}\\\\C = \frac{1}{8 \times 10^{24} \;\times \;15 \times 10^{-3} } \\\\C = \frac{1}{1.2 \times 10^{23}} \\\\C= 8.34 \times 10^{-24} \;F[/tex]
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Which of the following statement regarding orbits is true?
A) In a orbit, the satellite and the central body are the two Foci of the eclipse.
B) the orbit of the planet has an elliptical shape with the Sun at one Focus.
C) an elliptical orbit there is one focus and the satellite is located there.
D) the sun and a planet are the two Foci of an orbit
The correct option is option (B)
The orbit of the planets is elliptical with the sun at one of the foci.
Orbit of planets:According to Kepler's Laws:
The orbit of a planet around the Sun is an ellipse, with the Sun in one of the focal points of that ellipse. The planet's orbit lies in a plane, called the orbital plane. The point on the orbit closest to the attracting body is the periapsis. The point farthest from the attracting body is called the apoapsis. As the planet moves in its orbit, the line from the Sun to the planet sweeps a constant area of the orbital plane for a given period of time. This means that the planet moves faster near its perihelion than near its aphelion, because at the smaller distance it needs to trace a greater arc to cover the same area. This law is usually stated as equal areas in equal time.For a given orbit, the ratio of the cube of its semi-major axis to the square of its period is constant.Learn more about Kepler's Laws:
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You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
352 m/s
Explanation:
The velocity of sound in air is approximated as:
v ≈ 331 + 0.6 T
where v is the velocity in m/s and T is the temperature in Celsius.
At T = 35:
v = 331 + 0.6 (35)
v = 352 m/s
1/1
2. Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the
volume flow rate of the fluid in m^3/s *
Answer:
9.42*10^-4 m^3/s
Explanation:
d=3/100 m
=0.03 m
A=3.14*0.03^2/6
=4.71*10^-4 m^2
Volume flow rate V = A * s
= 4.71*10^-4 * 2
= 9.42*10^-4
So, the volume flw rate of fluid is 9.42*10^-4 m^3/s
A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.
Answer:
The time rate of change of the electric field between the plates is [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 2.8 \ cm = 0.028 \ m[/tex]
The distance of separation is [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]
The current is [tex]I = 5 \ A[/tex]
Generally the electric field generated is mathematically represented as
[tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value
[tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So the time rate of change of the electric field between the plates is mathematically represented as
[tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]
But [tex]\frac{q}{t } = I[/tex]
So
[tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]
substituting values
[tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]
[tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
[tex]B = \frac{\mu_o NI}{l} \\\\[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T[/tex]
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?
Answer:
(a) 0.59 ms
(b) 0.15 ms
(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.
Explanation:
(a) distance between ears = 20 cm = 0.2 m
speed of sound generated = 340 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{340}[/tex]
= 5.8824 × [tex]10^{-4}[/tex]
= 0.59 ms
The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.
(b) distance between ears = 20 cm = 0.2 m
speed of sound in water = 1530 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{1530}[/tex]
= 1.4815 × [tex]10^{-4}[/tex]
= 0.15 ms
The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.
(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.
A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds
B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds
C) The significance about the time interval of the two situations is that;
Transmission of sound varies with different mediums.
A) We are given;
Distance between the two ears; d = 20 cm = 0.2 m
Speed of sound; v = 340 m/s
Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;
t = d/v
t = 0.2/340
t = 0.588 × 10⁻³ seconds
B) We are now told that the speed of sound in water is 1530 m/s. Thus;
t = 0.2/1530
t = 0.131 × 10⁻⁵ seconds
C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.
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You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc
Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed
Answer:
Explanation:
d
g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision
Answer:
2
Explanation:
since the second object was in it stationary, it velocity is 0 m/s
EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4πrhoa4 rhoπa2 = a2 4 .
Answer:
Radius of gyration = a/2.
Explanation:
So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;
"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."
(NB: I changed the "rho" word to its symbol).
Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.
Therefore, the Radius of gyration = a/2.
Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.
Answer:
Explanation:
Using E= hc/wavelength
6.63 x10^-34 x3x10^8/ 236nm
19.86*10^-26/236*10^-9
=0.08*10^-35Joules
Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is:
Answer:
Three halves of a wavelength I.e 7lambda/2
Explanation:
See attached file pls
Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates the electric field field has a magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half-way between the plates
Answer:
Explanation:
Let the potential difference between the plate is V . Then in the first case
Electric field E between plate
E₁ = V / d
where d is separation between plate
When the plate separation becomes d / 2
Electric field E between plate
E₂ = V / d /2
= 2 V / d =2E₁
Or twice the earlier field
A bowling ball of mass 5 kg rolls off the edge of a building 20 meters tall. What is the work done by gravity during the fall, in Joules
Answer:
1000j
Explanation:
work done = force x distance
w = 5 x 10 x 20 = 1000joules
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-axis is the electric potential zero?
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, [tex]q_1=-3\ nC[/tex]
It is placed at a distance of 9 cm at x axis
Charge, [tex]q_2=+4\ nC[/tex]
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]
Here,
[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]
So,
[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]
Squaring both sides,
[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
According to the question,
Charge,
[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)Now,
→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]
or,
→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]
→ [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
here,
[tex]r_1 = \sqrt{y^2+81}[/tex]
[tex]r^2 = \sqrt{y^2+225}[/tex]
By substituting the values,
→ [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]
By applying cross-multiplication,
[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]
By squaring both sides, we get
→ [tex]9(y^2+225) = 16(y^2+81)[/tex]
[tex]9y^2+2025 = 16 y^2+1296[/tex]
[tex]2025-1296=7y^2[/tex]
[tex]7y^2=729[/tex]
[tex]y = 10.2 \ m[/tex]
Thus the solution above is correct.
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Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity:_____________.
(a) stays the same.
(b) Increases.
(c) Decreases.
(d) The answer would depend on the path of motion
Answer:
(b) Increases
Explanation:
The potential energy between two point charges is given as;
[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]
Where;
k is the coulomb's constant
q₁ ans q₂ are the two point charges
r is the distance between the two point charges
Since the two charges have opposite sign;
let q₁ be negative and q₂ be positive
Substitute in these charges we will have
[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]
The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.
Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 480 turns, its secondary 8 turns, and the input voltage is 118 V
Answer:
1.97 V
Explanation:
Applying
N1/N2 = V1/V2................... Equation 1
Where N1 = primary turns, N2 = Secondary turns, V1 = primary/input voltage, V2 = secondary/output voltage.
make V2 the subject of the equation
V2 = V1(N2/N1)............. Equation 2
Given: V1 = 118 V, N1 = 480 turns, V2 = 8 turns.
Substitute into equation 2
V2 = 118(8/480)
V2 = 1.97 V.
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness
This is note the complete question, the complete question is:
One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?
Answer: 3.04 D
Explanation:
when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.
An image of 0.47m from the eye will be ( 47 - 2 )
i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.
Also, the image distance will become negative
gap between lense and eye = 2cm
Therefore;
image distance d₁ = - 45cm = - 0.45m
object distance d₀ = 21 - 2 = 19cm = 0.19m
P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)
P = 1/f = 5.26315789 - 2.22222222
P = 1/f = 3.04093567 ≈ 3.04 D
A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?
Answer:
4.31 N
Explanation:
Given:
Δy = -55 m
v₀ = 0 m/s
v = -29 m/s
Find: a
v² = v₀² + 2aΔy
(-29 m/s)² = (0 m/s)² + 2a (-55 m)
a = -7.65 m/s²
Sum of forces in the y direction:
∑F = ma
R − mg = ma
R = m (g + a)
R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)
R = 4.31 N
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 2V0, what speed would it gain
Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀
An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner
Answer:
The rejected by the air conditioning system is 1.75 kilowatts.
Explanation:
A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:
[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]
Where:
[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.
[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.
[tex]\dot W[/tex] - Power input, measured in kilowatts.
The heat rejected is now cleared:
[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]
If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:
[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]
[tex]\dot Q_{H} = 1.75\,kW[/tex]
The rejected by the air conditioning system is 1.75 kilowatts.
You are given two infinite, parallel wires, each carrying current.The wires are separated by a distance, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
A. Is the force between the wires attractive orrepulsive?
B. What is the force per unit length between the two wires?
Answer:
A. Attractive
B. ( μ₀I² ) / ( 2πd )
Explanation:
A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.
B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -
[tex]B_1[/tex] = μ₀I / 2πd
The force experienced by wire 2 should thus be -
[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )
= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )
= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )
Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...
( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution
Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now
Answer:
F₂ = 1/3 FExplanation:
Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.
Mathematically, F = GMaMb/R² ... 1
G is the gravitational constant
Ma and Mb are the masses of the balls
R is the distance between the balls
If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;
F₂ = GMa(3Mb)/(3R)²
F₂ = 3GMaMb/9R² ... 2
Dividing equation 1 by 2 we will have;
F₂/F = (3GMaMb/9R²)/GMaMb/R²
F₂/F = 3GMaMb/9R² * GMaMb/R²
F₂/F = 3/9
F₂/F = 1/3
F₂ = 1/3 F
This shows that the magnitude of the new attractive force is one-third that of the initial attractive force
Shelly experiences a backward jolt when the driver starts the school bus. Which of the following explains this phenomenon?
Answer:
A. the inertia of shelly
Explanation:
Answer:
A. the inertia of Shelly
Explanation:
newtons first law states that an object at rest stays at rest unless on opposite force act against it, so when the buss started to move it acted on shelly forcing her to move back. Shelly's inertia was at a halt and was forced back by the motion of the bus.
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
The distance measured between five successive crests of a wave motion executed by a photon of an electromagnetic radiation is 2.4cm.
1. Determine the wavelength in micrometers
2. what is the frequency of the photon
3. what is the energy possessed by the photon.
Answer:
1. λ = 0.48 cm = 4800 μm
2. υ = 6.25 x 10¹⁰ Hz
3. E = 4.14 x 10⁻²³ J
Explanation:
1.
Since, the wavelength is defined as the distance between two consecutive or successive crests or troughs. Therefore, in this case the wavelength will be equal to:
Wavelength = λ = Distance between 5 successive crests/5
λ = 2.4 cm/5
λ = 0.48 cm = 4800 μm
2.
The frequency of photon can be given as:
υ = c/λ
where,
υ = frequency of photon = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 0.48 cm = 0.0048 m
Therefore,
υ = (3 x 10⁸ m/s)/(0.0048 m)
υ = 6.25 x 10¹⁰ Hz
3.
Now, the energy of photon is given as:
E = hυ
where,
E = Energy = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(6.25 x 10¹⁰ Hz)
E = 4.14 x 10⁻²³ J