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Vlad1618 [11]
2 years ago
11

A 4kg block sitting on the floor, how much potential energy does it have?

Physics
2 answers:
aniked [119]2 years ago
8 0
P.E.=mgh

P.E.= mass×gravity×height, therefore 4kg×10×0= 0 J

K.E. is also equal to 0 because the block is not moving, is in the state of Inertia
prohojiy [21]2 years ago
3 0

Well, there you have a very important principle wrapped up in that question.

There's actually no such thing as a real, actual amount of potential energy.
There's only potential <em><u>relative to some place</u></em>.  It's the work you have to do
to lift the object from that reference place to wherever it is now.  It's also
the kinetic energy the object would have if it fell down to the reference place
from where it is now.

Here's the formula for potential energy:    PE = (mass) x (gravity) x (<em><u>height</u></em><u>)</u> .

So naturally, when you use that formula, you need to decide "height above what ?"

If you're reading a book while you're flying in a passenger jet, the book's PE is
(M x G x 0 meters) relative to your lap, (M x G x 1 meter) relative to the floor of the
plane, (M x G x 10,000 meters) relative to the ground, and maybe (M x G x 25,000 meters)
relative to the bottom of the ocean.

Let's say that gravity is 9.8 m/s² .

Then a 4kg block sitting on the floor has (39.2 x 0 meters) PE relative to the floor
it's sitting on, also (39.2 x 3 meters) relative to the floor that's one floor downstairs,
also (39.2 x 30 meters) relative to 10 floors downstairs, and if it's on the top floor of
the Amoco/Aon Center in Chicago, maybe (39.2 x 345 meters) relative to the floor
in the coffee shop that's off the lobby on the ground floor. 

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

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3 years ago
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kenny6666 [7]

Answer:

C to/tc

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2 years ago
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Read 2 more answers
Spaceship A is 10 meters long and approaching you from the south at a speed of.7c while spaceship B, which is also 10 meters lon
Valentin [98]

Answer:

a) 0.94 C

b) 7.14 m

c) 3.11 m

d) 1.40 s

e) 2.93 s

Explanation:

First we need to set up a coordinate system. This will have the positive X axis pointing north. So spaceship A has positive speed, and spaceship B has negative speed.

The Lorentz transformation for speed is:

u' = \frac{u - v}{1 - \frac{u*v}{c^2}}

u: speed of spaceship A as observed by you

v: speed of spaceship B as observed by you

In the case of the speed of spaceship A as observed by spaceship B:

u' = \frac{0.7c - (-0.7c)}{1 - \frac{0.7c*(-0.7c)}{c^2}} = 0.94c

The transform for lengths is:

L = L0 * \sqrt{1 - \frac{v^2}{c^2}}

For the case of spaceship A as observed by you:

L = 10 m * \sqrt{1 - \frac{(0.7c)^2}{c^2}} = 7.14 m

For the case of spaceship A as observed by spaceship B:

L = 10 m * \sqrt{1 - \frac{(0.94c)^2}{c^2}} = 3.12 m

The time dilation equation is:

T = \frac{T0}{\sqrt{1-\frac{v^2}{c^2}}}

For the case of the event as observed by you:

\frac{1 s}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = 1.40 s

For the case of the event as observed by spaceship B:

\frac{1 s}{\sqrt{1-\frac{(0.94c)^2}{c^2}}} = 2.93 s

3 0
3 years ago
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