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RSB [31]
3 years ago
5

Samuel hits a ball, which is initially at rest, off of a tee. What is the velocity of the ball after it is hit, if Samuel applie

s 30 J of work to the 1.1 kg ball? It will be helpful to show your work on a separate sheet of paper.
Physics
1 answer:
aniked [119]3 years ago
8 0

Answer:

7.39 m/s

Explanation:

Applying

K.E = 1/2mv²..................... Equation 1

Where K.E = Kinetic Energy, m = mass of the ball, v = velocity of the ball.

Make v the subject of the equation

v = √(2K.E/m)................. Euqation 2

From the question,

Given: K.E = 30 J, m = 1.1kg

Substitute these values into equation 2

v = √(2×30/1.1)

v = √54.54

v = 7.39 m/s

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2 years ago
What is the magnitude of a vector with components (15 m, 8 m)?
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3 years ago
Read 2 more answers
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
3 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
What is the scientific revolution??
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