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Ymorist [56]
3 years ago
7

A 5 μF capacitor is connected to a 12 V battery. The charge on each plate of the capacitor is:

Physics
1 answer:
ziro4ka [17]3 years ago
3 0
1 farad = 1 coulomb/volt

5 μF = 5 x 10⁻⁶ coulomb/volt

        = 60 x 10⁻⁶ coulomb / 12 volts

The charge is 60 x 10⁻⁶ coulombs = 6 x 10⁻⁵  (choice-A) 


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We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
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-4.905t^2+11.1t-0.31=0
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3 years ago
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alexdok [17]

Answer:

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jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

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Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

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Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

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Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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