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drek231 [11]
3 years ago
9

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

enth.
3 kg: ____kJ

6 kg: ____kJ

9 kg: ____kJ

Physics
1 answer:
Goryan [66]3 years ago
6 0

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.  </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy U_{p} will be:  

U_{p}=mgh  

Where m is the mass of the object, g=9.8m/s^{2} the acceleration due gravity and h=500m the height of the object.  

Knowing this, let's begin with the calculaations:

For m=3kg

U_{p}=(3kg)(9.8m/s^{2})(500m)  

U_{p}=14700J=14.7kJ  

For m=6kg

U_{p}=(6kg)(9.8m/s^{2})(500m)  

U_{p}=29400J=29.4kJ  

For m=9kg

U_{p}=(9kg)(9.8m/s^{2})(500m)  

U_{p}=44100J=44.1kJ  

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2 years ago
In an experiment, an object is heated.
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Answer:

B) 300 J/°C

Explanation:

Q = 3kJ = 3000J

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3 years ago
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?
Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u =  6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0
3 years ago
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0
3 years ago
A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l
katen-ka-za [31]

Answer:

v'_2=3V

Explanation:

From the question we are told that:

Bowling ball Speed v_1=5 m/s

Beach ball Speed v_2=2 m/s

Let The Mass be equal i.e

 M_1=M_2

Therefore

Generally the equation for Velocity of beach ball after collision v'_2 is mathematically given by

Since Velocity is Vector Quantity

Therefore

 v'_2=v_1-v_2

 v'_2=5-2

 v'_2=3V

3 0
3 years ago
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