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Juliette [100K]
3 years ago
6

I need help please .....

Physics
1 answer:
arsen [322]3 years ago
3 0

Answer:

Explanation:

All of the energy in the pendulum is kinetic energy and there is no gravitational potential energy. However, the total energy is constant as a function of time.

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Suppose all of the apples have a mass of 0.10 kg. If apple (A) is sitting on the ground, what is its potential energy?
nekit [7.7K]

Answer:

The answer is C 0.0

Explanation:

Because its sitting still on the ground.

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3 years ago
Can Anyone help me with this?
agasfer [191]
1. Velocity

2. Time

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4 years ago
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a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

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3 years ago
Help with electrician homework
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Answer:

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5 0
3 years ago
Calculate the potential energy of a 1200 kg boulder on a cliff 45 m above the ground. ​
Softa [21]

Explanation:

Given mass of the object is 1200kg and it's placed at a height of 45m above the ground. As we know that potential energy is ,

\longrightarrow PE = mgh

where,

  • m is the mass of the body .
  • g is acceleration due to gravity.
  • h is the height above the ground .

Substituting the respective values ,

\longrightarrow P.E. = 1200kg * 10m/s^2* 45m

Multiply ,

\longrightarrow P.E. = 540000J

<u>Hence</u><u> the</u><u> </u><u>potential</u><u> energy</u><u> is</u><u> </u><u>5</u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>J</u><u> </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em>.</em>

4 0
3 years ago
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