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Zarrin [17]
4 years ago
12

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?
Physics
1 answer:
AlladinOne [14]4 years ago
7 0

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight W = 150 N

Length of rope r = 4.1 m

Initial speed of ball v = 5.5 \frac{m}{s}

For finding the tension in the rope,

First find the mass of rod,

mg = 150                          ( g = 9.8 \frac{m}{s^{2} } )

  m = \frac{150}{9.8}

  m = 15.3 kg

Tension in the rope is,

  T = mg + \frac{mv^{2} }{r}

  T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}

  T = 262.88 N

Therefore, the tension in the rope is 262.88 N

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How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10
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Answer:

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Work done would be done by

W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J

The work required to stop the electron is -2.00876\times 10^{-18}\ J

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3 years ago
A skateboarder jumps horizontally off the top of a staircase and lands at bottom of the stairs. The staircase has a horizontal l
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Answer:

The vertical velocity of the skater upon landing is 10.788 meters per second.

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v - Final vertical speed, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

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v = 0\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (1.10\,s)

v = -10.788\,\frac{m}{s}

The vertical velocity of the skater upon landing is 10.788 meters per second.

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