PHYSICS COLLEGE

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 5.8 kV potential distance d = 2.8 mm ? Approximate the area of a shoe as 30 cm x 8 cm. Express your answer using two significant figures.

Answers

Answer 1

Answer:

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$(Q)/(V)=(A\epsilon_0)/(d)\n\Rightarrow \Q=V(A\epsilon_0)/(d)\n\Rightarrow Q=5.8* 10^3(0.024* 8.85* 10^(-12))/(0.0028)\n\Rightarrow Q=4.4* 10^(-7)\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

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A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound

Answers

Answer:

about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600 seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

Answers

Answer:

$X - Xo = 54m$

k = 1/18

Explanation:

Data:

a = -k $t^(2)$ $(m)/(s^(2) )$

to = 0s Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

$a = (dV)/(dT)$

rearranging...

$a*dT = dV$

Then, we must integrate both sides:

$\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^(2) } \, dT$

$V - Vo = -k(t^(3) )/(3)$

V = 0 because the exercise says that the car change it's direction:

$0 - 12 = -k(6^(3) )/(3)$

k = 1/6

In order to find X - Xo we must integer v*dT = dX

$V - Vo = -k(t^(3) )/(3)$

so...

$(Vo -k(t^(3) )/(3))dT = dX$

$\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k(t^(3) )/(3) } \, dT$

integrating...

$X - Xo = Vot -k(t^(4) )/(12)$

$X - Xo = 12*6 -(1)/(6)* (6^(4) )/(12)$

X - Xo = 54m

What kind of energy is produce when sun reaches solar panel?

Answers

Answer:

Radient to ElEcTrIcAAl

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

The energy produced when the sun reaches solar panel is nuclear fusion

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