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Likurg_2 [28]
3 years ago
8

A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is conne

cted to a movable object. The spring and object are compressed by 0.068 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 17.8 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.034 m relative to its unstrained length
Physics
1 answer:
JulsSmile [24]3 years ago
3 0

Answer:

v_f = 1.05 m/s

Explanation:

From conservation of energy;

E_f = E_i

Thus,

(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²

This reduces to;

(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²

Making v_f the subject, we have;

v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]

We know that ω = √(k/m)

Thus,

v_f = ω[√((x_i)² - (x_f)²)]

Plugging in the relevant values to obtain;

v_f = 17.8[√((0.068)² - (0.034)²)]

v_f = 17.8[0.059] = 1.05 m/s

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Answer:

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Explanation:

I hope I helped and I hope u get what your looking for!!

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An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is
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Calculate the change in internal energy (δe) for a system that is giving off 25.0 kj of heat and is changing from 12.00 l to 6.0
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Since the system itself is giving off heat, this is a reduction in the internal energy.

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Since work is being done on the system, therefore it is an additional energy to the system. Work is given as:

work = - P dV

work = - 1.50 atm (6 L – 12 L)

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Since it is given that 1 L atm is equivalent to 101.3 J, therefore the total energy added is:

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<span>Therefore, approximately 24.1 kJ of energy is lost by the system in the total process.</span>

<span>
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<span>Answer:</span>

<span>-24.1 kJ</span>

8 0
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