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tia_tia [17]
4 years ago
11

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

eased from rest at a vertical position that is not the system’s equilibrium position. The system oscillates near Earth’s surface. The system is then taken to the Moon’s surface, where the gravitational field strength is nearly 16 that of the gravitational field strength near Earth's surface. Which of the following claims is correct about the period of oscillation for the system?
Physics
1 answer:
vodka [1.7K]4 years ago
4 0

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

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You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club an
Anna71 [15]

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          Em_{f} = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

6 0
3 years ago
Please help on this one?
DedPeter [7]

The answer is C. Hope this helps.

3 0
3 years ago
Read 2 more answers
What is a creative name for an element? (like oxygen, phosphorus) try to make it good
Inga [223]
Oxyles i think this suite Oxyles what do you think.
4 0
3 years ago
A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0
irakobra [83]
<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend 40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ . in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , v=\dfrac{total\ distance }{total\ times}

v=\dfrac{\dfrac{10}{60}\times 50+\dfrac{19}{60}\times 100+\dfrac{60}{60}\times 55}{\dfrac{10}{60}+\dfrac{10}{60}+\dfrac{60}{60}+ \dfrac{40}{60}}\\\\\\v=47.5\ m/s

Hence, this is the required solution.

Learn More :

Average speed

https://brainly.in/question/12701198

7 0
3 years ago
An 89 kg person climbs up a uniform 13 kg ladder. The ladder is 5.9 m long; its lower end rests on a rough horizontal floor (sta
lapo4ka [179]

Answer:

The maximum distance the person will reach before he slip is 1.82m.

Not even close to the middle of the ladder.

Explanation:

Check attachment for solution

8 0
3 years ago
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