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Irina-Kira [14]
3 years ago
11

You compress a spring by a certain distance. then you decide to compress it further so that its elastic potential energy increas

es by another 50%. what was the percent increase in the spring's compression distance?
Physics
1 answer:
Likurg_2 [28]3 years ago
5 0
Elastic potential energy = 1/2 k * change of x^2
k- coefficient
x - change in length.
to increase energy 1.5 times you have to change x (compress) into \sqrt{1.5}  times (it's abot 1.22 or 22%)
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<span>Vf = Vi + at
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3 years ago
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boyakko [2]

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Both objects have the same magnitude of momentum.

Explanation:

If an object of mass m is moving at a velocity of v, the momentum of that object would be m\, v.

The 100\; {\rm g} (0.1\; {\rm kg}) object is moving at a speed of 1\; {\rm m\cdot s^{-1}}. The magnitude of the momentum of this object would be 0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

Similarly, the momentum of the 1\; {\rm g} (10^{-3}\; {\rm kg}) object moving at a speed of 100\; {\rm m\cdot s^{-1}} would be 10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

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7 0
1 year ago
The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ
olya-2409 [2.1K]

Answer: 8400 J

Explanation:

The formula referenced in the question is:

Q=m. c. \Delta T  

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c=4200 \frac{J}{kg\°C}  is the specific heat capacity of  water

\Delta T=20\°C  is the variation in temperature

Solving:

Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)  

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2 years ago
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Lunna [17]

Answer:

A.

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I think it might be the big number A

8 0
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