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PtichkaEL [24]
3 years ago
12

#4 HELP PLEASE! Find Gravitational Potential Energy! THANK YOU!

Physics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

\Delta GPE=-3.92\ 10^9\ J

Explanation:

<u>Gravitational Potential Energy</u>

It's the capacity of an object to do work due to its relative height from a fixed reference point.

It's computed as

GPE=m.g.h

Where m is the mass of the object, h is its height and g is the acceleration of gravity, g=9.8 m/sec^2

The mass of water is given as

m=8,000,000\ kg

The height above the rocks is h=50 m. Let's compute the GPE

GPE_1=(8,000,000\ kg)(9.8\ m/s^2)(50\ m)

GPE_1=3,920,000,000\ Joule

It should be expressed in scientific notation

GPE_1=3.92\ 10^9\ J

The GPE at the bottom, where h=0

GPE_2=0

The change of gravitational potential energy is:

\Delta GPE=GPE_2-GPE_1

\boxed{\Delta GPE=-3.92\ 10^9\ J}

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A coil is connected in series with a 8.41 kΩ resistor. An ideal 68.6 V battery is applied across the two devices, and the curren
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Answer:

The inductance and stored energy are 234 H and 3.15\times10^{-4}\ J

Explanation:

Given that,

Resistance R= 8.41 kΩ

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Current I = 1.64 mA

Time t = 6.26 ms

We need to calculate the maximum current of the coil

Using formula of maximum current

I_{max}=\dfrac{V}{R}

I_{max}=\dfrac{68.6}{8.41\times10^{3}}

I_{max}=8.15\times10^{-3}\ A

We need to calculate the inductance of the coil

I_{f}=I(1-e^{\frac{-t}{\tau}})

t=-\dfrac{L}{R}In(1-\dfrac{I_{f}}{I_{max}})

6.26\times10^{-3}=-\dfrac{L}{8.41\times10^{3}}In(1-\dfrac{1.64\times10^{-3}}{8.15\times10^{-3}})

L=\dfrac{{6.26\times10^{-3}\times8.41\times10^{3}}}{ln(1-\dfrac{1.64\times10^{-3}}{8.15\times10^{-3}})}

L=234\ H

(b). We need to calculate the stored energy

Using formula of stored energy

U=\dfrac{1}{2}LI^2

U=\dfrac{1}{2}\times234\times(1.64\times10^{-3})^2

U=3.15\times10^{-4}\ J

Hence, The inductance and stored energy are 234 H and 3.15\times10^{-4}\ J

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3 years ago
A plane startling from rest accelerates at 3 m/s Square. Calculate the increase in velocity after 25 sec
Tatiana [17]

Answer:

75 m/s

Explanation:

We can apply motion equations here

V = U + a * t

V =  velocity @ t time

U = initial velocity

a = acceleration

t = time taken

V = U + a * t

V = 0+ 3 * 25

V = 75 m/s

After  25 seconds , subjected to the given acceleration velocity is increased from 0 to 75 m/s

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3 years ago
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