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Blababa [14]
3 years ago
9

A 513 g ball strikes a wall at 14.7 m/s and rebounds at 11.3 m/s. The ball is in contact with the wall for 0.038 s. What is the

magnitude of the average force acting on the ball during the collision? Answer in units of N.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

consider the velocity of the ball towards the wall as negative and away from the wall as positive.

m = mass of the ball = 513 g = 0.513 kg

v₀ =  initial velocity of the ball towards the wall before collision = - 14.7 m/s

v = final velocity of the ball away from the wall after collision = 11.3 m/s

t = time of contact with the wall = 0.038 sec

F = average force acting on the ball

using impulse-change in momentum equation , average force is given as

F = m (v - v₀)/t

inserting the values

F = (0.513) (11.3 - (- 14.7))/0.038

F = 351 N


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2 years ago
A certain superconducting magnet in the form of a solenoid of length 0.56 m can generate a magnetic field of 6.5 T in its core w
Gnom [1K]

Answer:

The value  is N =36203 \  turns

Explanation:

From the question we are told that

   The length of the solenoid is  l = 0.56 \  m

   The magnetic field is B  =  6.5 \ T

    The current is I = 80 \ A

     The desired temperature is  T = 4.2 \ K

Generally the magnetic field is mathematically represented as

       B  = \frac{\mu_o * N * I }{L }

=>     N = \frac{B  * L }{\mu_o * I }

Here  \mu_o is the permeability of free space with value  

      \mu_o =  4\pi * 10^{-7} N/A^2

So

     N = \frac{6.5  * 0.56 }{ 4\pi * 10^{-7} *  80  }

=>   N =36203 \  turns

6 0
3 years ago
An object starts at the 100 m mark and ends up at the -100 m mark, 50
horsena [70]
It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)
8 0
2 years ago
Five scientist who travelled to space​
Masja [62]

Answer:

The Most Famous Astronomers of All Time. Karl Tate, SPACE.com. ...

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6 0
3 years ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0
3 years ago
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