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belka [17]
3 years ago
7

A rubber ball with a mass of 0.30 kg is dropped onto a steel plate. The ball's velocity just before impact is 4.5 m/s and just a

fter impact is 4.2 m/s and just after impact is 4.2 m/s. What is the change in the ball's momentum?
Physics
1 answer:
My name is Ann [436]3 years ago
3 0

Answer:

Change in momentum will be -2.61 kgm/sec

Explanation:

We have given mass of the rubber ball m = 0.30 kg

Velocity of the ball before the impact v_1=4.5m/sec

Velocity of ball after impact v_2=-4.2m/sec ( negative sign is due to opposite direction of motion )

Change in momentum is given by m(v_2-v_1)=0.3\times (-4.2-4.5)=0.3\times =0.3\times -8.7=-2.61kgm/sec ( negative sign shows the direction of change in momentum )

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Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it
makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

 Here, we have been given the height through which the ball drops;

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The right motion equation is;

      V²  = U² + 2gh

V is the final velocity

U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

       V² = 0² + 2 x 9.8 x 12.6

      V²  = 246.96

       V = √246.96  = 15.7m/s

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How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of
Mekhanik [1.2K]

Answer:

# _units = 1000

Explanation:

This exercise we can use a direct proportion rule.

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the volume of a body of radius r = 10

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Read 2 more answers
The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 k
LuckyWell [14K]

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

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I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3)(90)(4.44)^2 = 0.33 x 90 x 19.7136 = 585.49

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KE = 0.00089 J

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I = (1/3)(58.2)(2.79)^2 = 0.33 x 58.2 x 2.79^2 = 149.5

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KE = 1/2(1/3)(90)(4.44)^2(pi/21600)^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J

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4 years ago
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