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rewona [7]
3 years ago
5

Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp

er sheet carries +5.00 µC/m2. The electric field midway between the sheets is 4.25 × 105 N/C pointing downward. What is the surface charge density on the lower sheet? (ε0 = 8.85 × 10-12 C2/N · m2)
Physics
1 answer:
sashaice [31]3 years ago
8 0

This problem refers to a parallel plate capacitor. There is an electric field between the two plates. The working equation to be used is the Gauss’s Law which is

Electric field = Surface charge density / ε0

The answer is -2.52 μC/m2.

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You have designed and constructed a solenoid to produce a magnetic field equal in magnitude to that of the Earth (5.0 10-5 T). I
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Answer:

I = 0.03637 A

Explanation:

The given data in the question is

Magnetic field : B = 5.0 \times 10^{-5} T

Turns : N = 350

Length : L = 32 cm = 0.32 m

So, number of turns per unit length :

n=\frac{N}{L}

n=\frac{350\: turns}{0.32 m}

n=1093.75 \: turns \: per \: meter

If current is I , then magnetic field is given by

B = \mu_{0} \times n \times I

Also,

I = \frac{B}{\mu_{0} \times n}

Insert the values

I = \frac{5.0 \times 10^{-5}}{4 \pi \times 10^{-7} \times 1093.75} A

I = 0.03637 A

The current will be I = 0.03637 A

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3 years ago
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a
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Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

v = \sqrt{\frac{GM}{R+h}}

here we know that

M = 5.98 \times 10^{24} kg

R = 6.37 \times 10^6 m

h = 900 km = 9.0 \times 10^5 m

now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

v = 7407.1 m/s

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

v_{rel} = \sqrt{2} v

v_{rel} = 1.05 \times 10^4 m/s

6 0
3 years ago
When the compressor of an air conditioner starts up, it draws a current of 50 A. If the start-up time is 0.60 s, then the amount
7nadin3 [17]

Answer:

30 C

Explanation:

Given:

Current flowing in the circuit (I) = 50 A

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Now, we know that, charge drawn in through a cross sectional area of the circuit is given as:

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q=50\ A\times 0.60\ s\\\\q=30\ C

Therefore, the amount of charge drawn in the circuit at the start-up of the compressor of an air conditioner is 30 C.

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