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2 years ago

A rocket has a mass of 156,789 kg and is traveling at 45.6 m/s. How much kinetic energy does the rocket

Physics

1 answer:

Flura [38]2 years ago

7 0

Explanation:

K.E =1/2 mv^2

=1/2(156789)(45.6)^2

=78,394.5 × 2,079.36

=163,010,387.52 kg m/s

This should be your answer.

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For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or

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Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

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2 years ago

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Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can

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The answer is a in the center of active volcanos

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2 years ago

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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3 years ago

In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will

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