Answer:
Total time taken(T) = 1.1 hour
Distance = 75.68 km
Explanation:
Given:
Average speed = 68.8 km/h
Constant speed = 98.5 km/h
Rest time = 20 min = 20 / 60 = 0.3333 hour
Find:
Total time taken(T)
Total distance (D)
Computation:
Distance = speed × time
D = 68.8 × t.........Eq1
and
D = 98.5 × [t-0.33]
D = 98.5 t - 32.8333.........Eq2
From Eq1 and Eq2
68.8 t = 98.5 t - 32.83333
29.7 t = 32.83333
t = 1.1
Total time taken(T) = 1.1 hour
Distance = speed × time
Distance = 68.8 × 1.1
Distance = 75.68 km
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.
A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward
Answer:
The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s
Explanation:
Given;
speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s
speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s
Determine the speed of the ball relative to the ground if the ball is thrown forward.
If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.
[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]
Therefore, the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.
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.
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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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
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:
Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.
Required:
What is the maximum emf induced in the loop?
Answer:
The maximum emf induced in the loop is 0.132 Volts
Explanation:
Given;
radius of the circular loop, r = 9.5 cm
intensity of the wave, I = 0.0295 W/m²
wavelength, λ = 6.40 m
The intensity of the wave is given as;
[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]
where;
B₀ is the amplitude of the field
c is the speed of light = 3 x 10⁸ m/s
μ₀ is permeability of free space = 4π x 10⁻⁵ m/A
[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]
Area of the circular loop;
A = πr²
A = π(0.095)²
A = 0.0284 m²
Frequency of the wave;
f = c / λ
f = (3 x 10⁸) / (6.4)
f = 46875000 Hz
Angular velocity of the wave;
ω = 2πf
ω = 2π(46875000)
ω = 294562500 rad/s
The maximum induced emf is calculated as;
emf = B₀Aω
= (1.572 x 10⁻⁸)(0.0284)(294562500)
= 0.132 Volts
Therefore, the maximum emf induced in the loop is 0.132 Volts
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
Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 232 and 194 newtons, forming an angle of 67
Answer:
408N at 89.89°
Explanation:
This problem requires that we resolve the force vectors into
x- and y
-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.
Finally, we will convert the resultant force into standard form and find the equilibrant.
Resolve into components:
F1x =F1cos 180°= 232(−1)=−232N
F1y=F1sin180°=0N
F2x=F2cos(−140°)=194(−0.766)=−148.6N
F1y=F1sin(−140°)=232(−0.643)=−149.17N
Note the change of the angle used to give the direction of
F2. Standard angles (rotation from thex
-axis; counterclockwise is +) should be used to avoid sign errors in the results.
Now, we add the components:
Fx=F1x+F2x=−380.6N
Fy=F1y+F1y=−148.17N
Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:
F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N
θ=tan−1(−148.17−380.6)
=89.89°
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 .
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
Express your answer numerically in kilowatts to two significant figures.
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Express the total force numerically, to two significant figures, in units of newtons.
Answer:
1) 14 kW
2) 4.67 x 10^-5 N
Explanation:
Area of solar panel = 10 m^2
Intensity of sun's radiation incident on earth = 1.4 kW/m^2
Solar power absorbed = ?
We know that the intensity of radiation on a given area is
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where I is the intensity of the radiation
P is the power absorbed due to this intensity on a given area
A is the area on which this radiation is incident
From the equation, we have
P = IA
P = 1.4 kW/m^2 x 10 m^2 = 14 kW
b) For a perfect absorbing surface, the radiation pressure is given as
p = I/c
where p is the radiation pressure
I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa
we know that Force = pressure x area
therefore force on the solar panels is
F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N
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
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.
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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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
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
A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is
Answer:
0.0081T
Explanation:
The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e
B ∝ I [tex]\frac{N}{L}[/tex]
B = μ₀ I [tex]\frac{N}{L}[/tex] ----------------(i)
Where;
μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²
Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by
L = 2π r
L = 2π (0.042)
L = 0.084π
The number of turns N = 1000
The current in the toroid = 1.7A
Substitute these values into equation (i) to get the magnetic field as follows;
B = 4π x 10⁻⁷ x 1.7 x [tex]\frac{1000}{0.084\pi }[/tex] [cancel out the πs and solve]
B = 0.0081T
The magnetic field along the central radius is 0.0081T
Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=
Answer:
Electric potential at position, x = -20 m, = -100 V
Electric potential at position, x = 0 m, = 0
Electric potential at position, x = 10 m, = 50 V
Electric potential at position, x = 11 m, = 55 V
Electric potential at position, x = 99 m, 495 V
Explanation:
Given;
magnitude of electric field, E = 5.0 V/m
at position x = 10 m, electric potential = 50 V
Electric potential at position, x = -20 m
V = Ex
V = 5 (-20)
V = -100 V
Electric potential at position, x = 0 m
V = Ex
V = 5(0)
V = 0
Electric potential at position, x = 10 m
V = Ex
V = 5(10)
V = 50 V
Electric potential at position, x = 11 m
V = Ex
V = 5(11)
V = 55 V
Electric potential at position, x = 99 m
V = Ex
V = 5(99)
V = 495 V
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids
Explanation:
It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.
As they are in equilibrium, the force of gravity due to each other is same. So,
[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]
So, the ratio of masses X/Y is 0.0847
As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.
Answer:
processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Explanation:
The temperature of planet Earth is due to two main types of process, internal and external.
Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect
External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.
These are the two terms that heat the Earth
In addition there are several processes so the planet loses energy,
* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission
* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system
* liquid (water) system that absorbs large amounts of heat to change state and temperature.
These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value
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
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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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
Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge
Answer:
sphere #1 carries positive charge and #2 carries negative charge
This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged
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]
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
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
A 4.7 kg solid sphere made of metal, whose density is 4000 kg /m3, is suspended by a cord. The density of water is 1000 kg/m3.When the sphere is immersed in water, the tension in the cord is closest to:
Answer:
34.58 N
Explanation:
From the question,
Tension in the cord = Weight of the solid metal- Upthrust
T = W-U................... Equation 1
But,
W = mg
Where m = mass of solid, g = acceleration due to gravity.
Given: m = 4.7 kg, g = 9.8 m/s²
W = 4.7(9.8) = 46.1 N.
U = (Density of water×Volume of water displaced.)gravity
U = D×V×g.
But,
V = Mass of solid/density of solid
V = 4.7/4000
V = 1.175×10⁻³ m³.
Therefore,
U = 1.175×10⁻³(1000)(9.8)
U = 11.52 N
Therefore,
T = 46.1-11.52
T = 34.58 N
Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. The amount of charge that passes per unit time is called
The amount of charge that passes per unit time is called electric current .
Current has dimensions of [Charge] / [Time] .
It's measured and described in units of ' Ampere ' .
1 Ampere means 1 Coulomb of charge passing a point every second.
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]