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katovenus [111]
3 years ago
14

A spring is stretched to a displacement of 3.4 m from equilibrium. Then the spring is released and allowed to recoil to a displa

cement of 1.9 m from equilibrium. The spring constant is 11 N/m. What best describes the work involved as the spring recoils?
Physics
1 answer:
Stells [14]3 years ago
5 0
Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
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Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

Acceleration = a = ?

F = ma

a = F/m

a = 2000/ 1000

a = 2 m/s^{2}

Q2: Solution:

Change in velocity = Δv = ?

acceleration = change in velocity / time

a = Δv/t

Δv = axt

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Δv = 20 m/s

Q3: Solution:

Impulse = I = ?

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I = 20000 Ns

Q4: Solution:

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Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

Δp = m x (vf-vi)

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vf = 20000/1000

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Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

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Δp = -20000 kgm/s

Q9: Solution:

Impulse = Δp

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Q10: Solution:

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F = -20000/4s

F = -5000 N

Q11: Solution:

F = ma

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Explanation:

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