Answer:
205 V
V[tex]_{R}[/tex] = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
[tex]V_{L} = - IwLsin(wt)[/tex]
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V[tex]_{R}[/tex] = IR
= (0.044 A) (93 Ω)
V[tex]_{R}[/tex] = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V[tex]_{R}[/tex] = 2.05 V
a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?
Answer:
8.78 AmpsExplanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
Hence the current when the unit/heater is operating is 8.78 AmpExplain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.
Answer:
Answer in explanation
Explanation:
Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.
The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.
Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.
This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.
If 1.7 kg of 238Pu is required to power the spacecraft's data transmitter, for how long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
The question is incomplete. Here is the complete question.
The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.
(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.
(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) 91 years.
Explanation:
(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:
Q(t) = [tex]0.5Q_{0}[/tex]
[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]
[tex]0.5=e^{-87.7k}[/tex]
[tex]ln(0.5)=ln(e^{-87.7k})[/tex]
[tex]ln(0.5)=-87.7k[/tex]
[tex]k = \frac{ln(0.5)}{-87.7}[/tex]
k = 0.0079
Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) Using function:
[tex]Q(t)=3.5e^{-0.0079t}[/tex]
[tex]1.7=3.5e^{-0.0079t}[/tex]
[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]
[tex]e^{-0.0079t}=0.4857[/tex]
[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]
[tex]-0.0079t=-0.7221[/tex]
[tex]t = \frac{-0.7221}{-0.0079}[/tex]
t = 91.41
t ≈ 91 years
Scientists will be able to receive data for approximately 91 years.
A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
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At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Answer:
B. mineral oil – density of 0.910 g/mL.
Explanation:
Hello,
In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.
In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.
Best regards.
A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green
The color that is reflected when a red card is illuminated by red light is white.
The color an object is perceived to have, depends on the frequency of light it reflects.
If white light incidents on a red filter, red is transmitted while blue and green are absorbed.
Consequently, when a red card is illuminated by red light, the red card will reflect back almost all the incident light on it, causing it to appear brighter which creates an illusion of white color to the eyes.
Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.
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A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Where should the mother sit to balance the seesaw?(1 point) at the opposite side of the seesaw on the end at the opposite side of the seesaw towards the middle on the same side of the seesaw towards the middle on the same end as her child
Answer:middle
Explanation:
Because it will make the seasaw balanced
There are two cells, one with OER as 2.5 and other as 7. Which cell is more sensitive to radiation?
1)The cell with OER 2.5
2)The Cell with OER 7
3)Both the cells
4)Insufficient data
Answer:
2)The Cell with OER 7
Explanation:
OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.
Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 μF capacitor from a camera flash unit retains a voltage of 140 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answer:
93.3x10^-3s
Explanation:
If
Resistance = 1.8 kΩ
Current = 50 mA
Capacitor = 120 μF
Voltage = 140 V
to calculate the discharge current
Applying the formula of discharge current
io=vo/R
io= 140/ 1.8x 10³
= 0.078A
to calculate the time
Applying the formula of current
io= vo/R e-t/RC
50= 140/1800e-t/RC
0.649= e-t/RC
-t/RC= ln( 0.649)
t = 0.432x 120x10^-6x 1800
t=93.3 x 10^-3seconds
A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?
Answer:
24
Explanation:
The mass = 3 kg
at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.
PE = mgh
but ME = PE + KE = constant (law of energy conservation)
KE is the kinetic energy
since KE is zero at this point, then,
ME = mgh
where m is the mass
g is acceleration due to gravity = 9.81 m/s^2
h is the height = O
ME = 3 x 9.81 x O
ME = 29.43-O
At point A the total ME is due to its PE and its kE
PE at this point = mgh = 3 x 9.81 x A = 29.43-A
KE = [tex]\frac{1}{2}mv^{2}[/tex]
velocity = v at this point, therefore,
KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]
therefore,
ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
Equating ME for the points O and A, we have
29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]
(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]
O - A = 0.051[tex]v^{2}[/tex] this is the distance between point O and A
For point B
PE = 29.43-B
KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex] (velocity is equal to [tex]5v[/tex] at this point)
therefore,
ME = 29.43-B + 37.5[tex]v^{2}[/tex]
Equating the ME for points A and B, we have
29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]
29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]
(A - B)29.43 = 36[tex]v^{2}[/tex]
A - B = 1.22[tex]v^{2}[/tex] this is the distance between points A and B
The distance between points A & B divided by the distance between points 0 & A will be
1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24
Describe at least two unique characteristics of the new space telescope.
Answer: Astro 1 will have a 10x larger field of view than the Hubble.
Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.
If you were to come back to our solar system in 6 billion years, what might you expect to find?
A) a red giant star
B) a rapidly spinning pulsar
C) a white dwarf
D) a black hole
E) Everything will be essentially the same as it is now
Answer:
A)a red giant star
1.
(a)
P
center
Figure 1
A ball is released at point P with a tangential velocity of 5 ms to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m? Give reasons. (4 marks]
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
A person starts at position zero, walks to position 8, then walks to position 5. Which answer correctly identifies the person's distance traveled? *
Answer:
Distance = 13 units
Explanation:
The overall path covered by an object during its journey is called distance covered.
In this problem, a person starts at position zero, walks to position 8, then walks to position 5.
We need to find the person's distance traveled. It can be calculated simply by adding all the positions i.e.
Distance = 0+8+5
Distance = 13
Hence, the distance covered by the person is 13 units.
Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.4 GHz, and as they do so they emit 2.4 GHz electromagnetic waves. What is the strength of the magnetic field?
Answer:
The magnetic field is 0.0857 T.
Explanation:
The electrons orbit the magnetic field with a centripetal force equal to
F = [tex]\frac{mv^{2} }{r}[/tex]
also, the force on an electron in a magnetic field is gotten as
F = Bqv
equating this two equations give
[tex]\frac{mv^{2} }{r}[/tex] = Bqv
mv/r = Bq
where m is the mass of the electron = 9.11 x 10^-31 kg
v is the the linear speed of the electron
B is the magnetic field on the electron
r is the radius of the orbital movement
q is the charge on an electron = 1.602 x 10^-19 C
but, the linear speed v = ωr
where ω is the angular speed of the electron
substituting into equation above, we have
mωr/r = Bq
which reduces to
mω = Bq
finally, w know that the angular speed is related to the frequency of the electron by
ω = 2πf
we then finally have
2mπf = Bq
where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz
substituting values into the equation, we have
2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19
B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T
= 85.7 mT
If the ac peak voltage across a 100-ohm resistor is 120 V, then the average power dissipated by the resistor is ________
Answer:
The average power dissipated is 72 W.
Explanation:
Given;
peak voltage of the AC circuit, V₀ = 120 V
resistance of the resistor, R = 100 -ohm
The average power dissipated by the resistor is given by;
[tex]P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}[/tex]
where;
[tex]V_{rms}[/tex] is the root-mean-square-voltage
[tex]V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V[/tex]
The average power dissipated by the resistor is calculated as;
[tex]P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W[/tex]
Therefore, the average power dissipated is 72 W.
Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential
Answer:
1. the sphere of the radius a
Explanation:
Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.
So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .
With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r
(for r ≥ R ) .
On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.
So Since b > a , the sphere of radius a will have the higher potential.
Also recall Because E = 0 inside a conductor, the potential
A football is kicked with a velocity of 18 m/s at an angle of 20°. What is the
ball's acceleration in the horizontal direction as it flies through the air?
Explanation:
It is given that,
The velocity of football is 18 m/s
It is projected at an angle of 20 degrees
We need to find the ball's acceleration in the horizontal direction as it flies through the air.
When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.
In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
An average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37∘C37 ∘ C, and the skin is typically 7C∘7C ∘ cooler. By how much does the person’s entropy change per second due to this heat transfer?
Answer:
-4.7 x 10^-3 J/K-s
Explanation:
The Power generated by metabolizing food = 80 W
The watt W is equivalent to the Joules per sec J/s
therefor power = 80 J/s
20% of this energy is not used for heating, amount available for heating is
==> H = 80% of 80 = 0.8 x 80 = 64 J/s
The inner body temperature = 37 °C = 273 + 37 = 310 K
The entropy of this inner body ΔS = ΔH/T
ΔS = 64/310 = 0.2065 J/K-s
The skin temperature is cooler than the inner body by 7 °C
Temperature of the skin = 37 - 7 = 30 °C = 273 + 30 = 303 K
The entropy of the skin = ΔS = ΔH/T
ΔS = 64/303 = 0.2112 J/K-s
change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)
==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is used?
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 1.5. Light incident parallel to the central axis is focused at a point _____________
Answer:
36cm from the surfaceExplanation:
Equation of refraction of a lens is expression according to the formula given below;
[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]
R is the radius of curvature of the convex refracting surface = 12cm
v is the image distance from the refracting surface
u is the object distance from the refracting surface
n₁ and n₂ are the refractive indices of air and the medium respectively
Given parameters
R = 12 cm
u = [tex]\infty[/tex] (since light incident is parallel to the axis)
n₁ = 1
n₂ = 1.5
Required
focus point of the light that is incident and parallel to the central axis (v)
Substituting this values into the given formula we will have;
[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]
Cross multiply
[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]
Hence Light incident parallel to the central axis is focused at a point 36cm from the surface
