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Juli2301 [7.4K]
3 years ago
6

An athlete with mass m running at speed v grabs a light rope that hangs from a ceiling of height H and swings to a maximum heigh

t of h1. In another room with a lower ceiling of height H/2, a second athlete with mass 2m running at the same speed v grabs a light rope hanging from the ceiling and swings to a maximum height of h2. How does the maximum height reached by the 2 athletes compare and why? (Pick one) a) The first athlete reaches a greater height, because the athlete swings on a longer rope; b) The second athlete reaches a greater height, because the athlete has a greater mass; c) The 2 athletes reach the same height, because the effect of the rope length offsets the effect of the athletes’ masses; d) The 2 athletes reach the same height, because the athletes run with the same speed.
Physics
1 answer:
-Dominant- [34]3 years ago
7 0

Answer:

d) The 2 athletes reach the same height, because the athletes run with the same speed.

Explanation:

In the whole process , kinetic energy is converted into potential energy .

1/2 m v² = mgh

v² = 2gh

h = v² / 2g

In this expression we see that height attained does not depend upon mass of the object . At the same time it also makes it clear that it depends upon velocity . As the velocity in both the cases are same , height attained by both of them will be same. Hence option d ) is correct.

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Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

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3 years ago
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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

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Here,

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

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Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

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We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

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