Your Answer Is in this attachment
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the car total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.
Answer:
45
Explanation:
ft/sec2
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25
Answer:
100Hz
Explanation:
In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz
Answer:
HZ 100 is the right answer hope you like it
Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks)
1. A flywheel begins rotating from rest, with an angular acceleration of 0.40 rad/s. a) What will its angular velocity be 3 seconds later? b) What angle will it have turned through in that time?
Answer:
(a) 1.2 rad/s
(b) 1.8 rad
Explanation:
Applying,
(a) α = (ω-ω')/t................ Equation 1
Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.
From the question,
Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)
Substitute these values into equation 1
0.40 = (ω-0)/3
ω = 0.4×3
ω = 1.2 rad/s
(b) Using,
∅ = ω't+αt²/2.................. Equation 2
Where ∅ = angle turned.
Substitutting the values above into equation 2
∅ = (0×3)+(0.4×3²)/2
∅ = 1.8 rad.
A typical radio frequency is 99.5 MHz, while a HD TV broadcast frequency is 600 MHz. Which one is your body position more likely to interfere with and why
I NEED YOUR HELP!!
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum area of rectangle
Computation :
Minimum area of rectangle = Length of rectangle x Width of rectangle
Minimum area of rectangle = 6 x 4
Minimum area of rectangle = 24 centimeter²
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the horizontal road is 0.735, how long will the skid marks be
Answer:
126.56 m
Explanation:
Applying,
-F = ma............. Equation 1
Where F = frictional force, m = mass of the car, a = acceleration.
Note: Frictional force is negative because it act in opposite direction to motion
But,
F = mgμ.......... Equation 2
Where g = acceleration due to gravity, μ = coefficient of friction
Substitute equation 2 in equation 1
-mgμ = ma
a = -gμ.............. Equation 3
From the question,
Given: μ = 0.735
Constant: 9.8 m/s²
Substitute these values in equation 3
a = -9.8×0.735
a = -7.203 m/s²
Finally,
Applying
v² = u²+2as.............. Equation 4
Where v = final velocity, u = initial velocity, s = distance
From the question,
Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²
Substitute these values into equation 4
0² = 42.7²+2(-7.203)s
-1823.29 = -14.406s
s = -1823.29/-14.406
s = 126.56 m
One ball is aprojected in the uapward directon with a certain velocity ‘v’ and other
is thrown downwards with the same velocity
Complete question is;
One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.
The ratio of their potential energies at highest points of their journey, will be:
Answer:
u² : (u cos θ)²
Explanation:
Maximum potential energy for the first ball will be at a maximum height of;
H = u²/2g
Thus;
PE = mg(u²/2g)
For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g
PE = mg((u cos θ)²/2g)
The ratios of the potential energies are;
mg(u²/2g) : mg((u cos θ)²/2g)
mg will will cancel out since they are of same mass.
Thus;
(u²/2g) : (u cos θ)²/2g
Again 2g will cancel out to give;
u² : (u cos θ)²
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V
Urgent please help me
1433 km
Explanation:
Let g' = the gravitational field strength at an altitude h
[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]
We also know that g at the earth's surface is
[tex]g = G\dfrac{M_E}{R_E^2}[/tex]
Since g' = (2/3)g, we can write
[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]
Simplifying the above expression by cancelling out common factors, we get
[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]
Taking the square root of both sides, this becomes
[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]
Solving for h, we get
[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]
[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]
What type of Literary Devices is this passage and provide an explanation
“A single flow’r he sent me, since we met.
All tenderly his messenger he chose;
Deep-hearted, pure, with scented dew still wet-
One perfect rose.”
By the admiring tone that the writer has for the gift that she/he received, it is clear that there's a lot of imagery. The writer also described the rose as "perfect", "scented dew still wet", and "pure", which further supports the idea that he/she is describing the gift.
Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house.
Required:
a. How many turns does the primary coil on the transformer have if the secondary coil has 130 turns?
b. No energy is lost in an ideal transformer, so the output power Pout from the secondary coil equals the input power Pin to the primary coil. Suppose a neighborhood transformer delivers 280 A at 120 V. What is the current in the 1.3×10^4 V line from the substation?
Answer:
a) N₁ = 14083 turns, b) I₁ = 2.58 A
Explanation:
The relationship that describes the relationship between the primary and secondary of the transformer is
[tex]\frac{V_2}{N_2} = \frac{V_1}{N_1}[/tex]
a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are
N₁ = [tex]N_2 \frac{V_1}{V_2}[/tex]
N₁ = [tex]130 \ \frac{13000}{120}[/tex]
N₁ = 14083 turns
b) since there are no losses, the power of the neighboring transformer is
P = V I
P = 120 280
P = 33600 W
this is the same power of the substation
P = V₁ I₁
I₁ = P / V₁
I₁ = 33600/13000
I₁ = 2.58 A
What is the meant of by renewable energy and non-renewrable with example of each.
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
Me Ayudan con este ejercicio por favor !!!
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
To appreciate the strength of gravity, calculate the number of steel cables that would be required, in the absence of gravity, to hold the Moon in orbit about the Earth. Each cable has a cross-sectional area of 1 square meter, and the tensile strength of steel is 370 MPa.
Answer:
[tex]N=5.36*10^{17}[/tex]
Explanation:
From the question we are told that:
Tensile strength of steel [tex]\sigma_c=370 MPa.=370*10^6[/tex]
Cross-sectional area [tex]A= 1m^2[/tex]
Generally the equation for Total Force is mathematically given by
[tex]F = \frac{G M m}{r^2}[/tex]
[tex]F=\frac{6.67*10^{-11}*7.35*10^{22}*5.97*10^{24}}{(384*10^3)^2}[/tex]
Since this is the force of each cable
[tex]F=\sigma_c NA[/tex]
[tex]F= 370*10^6*N*1[/tex]
Therefore
[tex]370*10^6*N*1=\frac{6.67*10^{-11}*7.35*10^{22}*5.97*10^{24}}{(384*10^3)^2}[/tex]
[tex]N=5.36*10^{17}[/tex]
A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Determine the average acceleration of the car during this time interval.
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 14 m/s
Finally, it comes to rest, v = 0
Time, t = 5.6 s
We need to find the average acceleration of the car during this time interval. We know that,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.
A wheel accelerates from rest to 20 rad/s at a uniform rate of 3.5 rad/s2. Through what angle (in radians) did the wheel turn while accelerating
Answer:
[tex]\theta=57.14rad[/tex]
Explanation:
From the question we are told that:
Angular Velocity [tex]\omega=20rad/s[/tex]
Acceleration [tex]a=3.5rads/s^2[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]\omega_2^2=\omega_1^2+2 a \theta[/tex]
Therefore
[tex]20^2=0+2*2.5*\theta[/tex]
[tex]\theta=\frac{400}{2*3.5}[/tex]
[tex]\theta=57.14rad[/tex]
Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes that Stephen’s displacement from the start line is 1500 m. Sarah says that she is incorrect and that his displacement from the start is actually 0 m. Which of the students is correct? Give reasoning for your answer.
Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
A 1500kg car start from rest and increases it velocity to 30mls in a time of 25sec. calculate the distance the car travel, how much force was use, how much work was done.
Answer:
workdone= 1/2mv^2
1/2×1500×30^2
675000J
A cylinder of volume 3 liter has Argon gas initially at 300 K, and 1.00 atm pressure. The piston compresses the gas to a new pressure of 3.60 atm. During this compression, the temperature is maintained constant by using an appropriate heat sink. Find (a) the final volume of the gas, (b) the work done on the gas by the piston, (c) the energy transferred out by heat.(d) If the process takes 20 milliseconds, what is the power
Answer:
Explanation:
Apply Boyle's law of gas
P₁ V₁ = P₂ V₂ where P₁ and P₂ are initial and final pressure , V₁ , V₂ are initial and final volume .
1.00 atm x 3 liter = 3.60 atm x V₂
V₂ = 0.833 liter.
b )
Work done on the gas by piston = 2.303 RT log P₂ / P₁
= 2.303 x 8.3 x 300 K x log 3.6 atm / 1 atm
= 3190 J .
c )
Q = ΔE + W
Q = 0 + 3190 J
Heat energy transferred out Q = 3190 J
d )
Power = work done / time
= 3190 J / .020 second
= 159500 W .
159.5 kW.
Answer:
(a) 0.833 L
(b) - 383.32J
(c) - 383.32 J
(d) 19166.2 W
Explanation:
initial volume, V = 3 L
initial temperature, T = 300 K
Initial pressure, P = 1 atm
final pressure, P' = 3.6 atm
Temperature is constant.
(a) Let the final volume is V'.
As the temperature is constant,
P V = P' V'
1 x 3 = 3.6 x V'
V' = 0.833 L
(b) Let the number of moles is n
[tex]P V = n R T \\\\1\times 1.01 \times 10^5\times 3\times 10^{-3} = n\times 8.31\times 300\\\\n = 0.12[/tex]
The work done in isothermal process is
[tex]W = n T T lon{\frac{V'}{V}}\\\\W = 0.12 \times 8.31\times 300\times ln {\frac{0.833}{3}}\\\\W = - 383.32 J[/tex]
(c) The energy is given by the first law of thermodynamics
dQ = dU + dW
Here, dU is the zero as the temperature is constant.
So, the heat energy is
dQ = dW = - 383.32 J
(d) Time, t = 20 ms
Power is
[tex]P = \frac{W}{t} \\\\P = \frac{383.32}{0.02} =19166.2 W[/tex]
heat from the sun comes on the earth by
Answer:
heat from the sun comes on the earth by radiation
In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first
Answer:
solid cylinder
Explanation:
the object that arrives first is the object that has more speed, let's use the concepts of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point
Em_f = K = ½ mv² + ½ I w²
since the body has rotational and translational movement
how energy is conserved
m g h = ½ mv² + ½ I w²
linear and angular velocity are related
v = w r
w = v / r
we substitute
m g h = ½ mv² + ½ I (v/r) ²
mg h = ½ v² (m + I /r²)
v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]
the tabulated moments of inertia for the bodies are
solid cylinder I = ½ m r²
hollow cylinder I = m r²
we look for the speed for each body
solid cylinder
v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]
v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]
let's call v₀ = [tex]\sqrt{2gh}[/tex]
v₁ = 0.816 v₀
hollow cylinder
v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]
v₂ = v₀ √½
v₂ = 0.707 v₀
Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive
A jet plane is speeding down the runway during takeoff. Air resistance is not negligible. Identify the forces on the jet.
Answer:
Weight , Drag , Thrust , Normal force
Explanation:
It is given that a jet plane is speeding during takeoff. The air resistance is negligible.
Therefore, the forces that acts on the jet plane are :
Weight
As the plane has some mass, the weight of the plane is acted upon it. The weight is acted in the downward direction.
Normal force or the Lift
The lift force is the force that helps the plane to move up in the air and it opposes the weight of the plane. It is the normal force to the weight of the plane and this force holds the plane in the air.
Thrust
The thrust force is the force which helps to move an aircraft in the direction of the motion. This force is created by the jet engine. This force moves the plane in the forward direction through the air.
Drag
Drag force is caused due to the difference in the velocity of the solid objects and the air. The drag force opposes the forward motion of the jet plane.
A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be
The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°
derive expression for pressure exerted by gas
A car driving down a road runs of gas and will eventually stop because of:
A. Friction
B. Thrust
C. It will remain in motion forever
OD. Gravity
