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Effectus [21]
3 years ago
7

Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and

the negative plate has a charge −q.The plate area is doubled, and the plate separation is reduced to half its initial separation.
a) What is the new charge on the negative plate?
Physics
1 answer:
dimaraw [331]3 years ago
8 0

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

c=\epsilon_0 \cdot \frac{A}{d}

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

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If pressurized air pressure is 350 kPa, atmospheric pressure is 100 kPa, initial atmospheric pressure is 100 kPa, initial accele
Travka [436]

Answer:

d=8.657mm

Explanation:

From the question we are told that

Pressurized air pressure is P_{air}=350 kPa,

Atmospheric pressure is P_a=100 kPa

Initial acceleration of the water rocket is a_i=0.5g.

Acceleration of the water rocket is  a_r=0.5g

Mass of water is M_w=0.5 kg

Generally total mass is given mathematically given as

T_M=0.5+0.5=>1kg

Generally the tension on the rocket is given mathematically given as

T=(P_{air}-P_a)A

T=(350-100) \frac{\pi d^2}{4}

T is also

T=\frac{3Mg}{2}

Therefore

T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}

T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}

d^2= \frac{3*1*9.81*4}{2(350-100) \pi}

d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}

d=8.657mm

therefore diameter of nozzle is mathematically given as

d=8.657mm

8 0
3 years ago
The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. a wire with a len
Annette [7]
The electrical resistance of the wire is proportional to its length L and inversely proportional to the square of its diameter, 1/r^2. 

The length of the wire in the problem is changed from 200 inches to 500 inches, so the new length is 2.5 times the initial length:
\frac{L_1}{L_0}= \frac{500}{200}=2.5
the diameter of the wire is not changed, so the new electrical resistance must be 2.5 times the original value:
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3 0
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HELP CHECK ANSWERS ONLY 3 QUESTIONS:
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Answer:

answer 1 is A, Answer 2 is A, Answer 3 is c

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The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. How much gravitational for
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Answer:

F = 29.9 N

Explanation:

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