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3 years ago

HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?

Physics

1 answer:

astra-53 [7]3 years ago

8 0

Given:-

Speed of the unicycle = 20 m/s
Time taken = 15 s

To Find: Distance travelled by the unicycle.

We know,

s = vt

where,

s = Distance travelled,
v = Speed &
t = Time taken.

Therefore,

s = (20 m/s)(15 s)

→ s = (20 m)(15)

→ s = 300 m (Ans.)

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Explanation:

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2 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

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Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

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The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

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$PE_{f} =0.5\ PE_{i}$

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2 years ago

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As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

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2 years ago

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3 years ago

HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?​

HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?