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Alex17521 [72]
3 years ago
8

A 5.0 V battery contains 775 C of charge. How much electricity energy can it produce?

Physics
1 answer:
KatRina [158]3 years ago
7 0

Answer:

3875J

Explanation:

Energy is defined as the power × time

And it's defined as

Power = IV - I- current and V- voltage

Now quantity of electricity; Q = I × t

Where I is current and t is time

Now Energy = I ×V×t = V× I×t = V× Q;

where Q is quantity of electricity 775C and V is 5.0volt

Hence 775 × (5) =3875J

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A chair at rest on the floor has two forces acting on it its own weight that pulls it downward and the floor pushing upward on the chair, both of these forces are acting on it but the net force is 0, so the chair remains at rest and its velocity stays at 0.

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A 2kg water balloon is flying at a rate of 4m/s^2. With what force will it hit its target?
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4. The flow of electric charge is know as
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electrons

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What energy transformation takes place in the<br> solar panel?
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3 years ago
Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
3 years ago
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