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GaryK [48]
3 years ago
7

A parallel-plate capacitor with plates of area 540 cm2 is charged to a potential difference V and is then disconnected from the

voltage source. When the plates are moved 0.6 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?
Physics
1 answer:
notsponge [240]3 years ago
5 0

Answer: 0.8 nC

Explanation: In order to explain this problem we have top use the expression for the capacitor of parallel plates, which is given by:

C=εo*A/d where A and d are the area and the separation of the plates.

Then we have

V=Q/C so ΔV= Q* Δd/(εo*A)

Q= 100* 0.054*8.85*10^-12/( 0.06)=0.8 nC

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Which device uses a rotating magnetic field to produce an electric current?
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Which of the following objects has the greatest density
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After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

5 0
2 years ago
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