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marishachu [46]
1 year ago
12

Problem Solving : Solve the following problem in a blank sheets of paper and submit your answer by uploading a jpeg or picture f

ile. five points will be given for a perfect computation. follow proper presentation of given data, cite formula and substitution used.
In a baseball game Daniel hit the ball with a force of 30 Newtons. This sent the ball flying through 300 meters in the midfield. How much force did Daniel do in the ball?


(show some solution)
Physics
1 answer:
crimeas [40]1 year ago
4 0

Daniel applies a 30 Newtons force on the ball that sends the ball 300 meters, which gives the work done by Daniel as 9 kJ

<h3>How can the work done be obtained from the given force and distance?</h3>

The given parameters are;

The force with which Daniel hit the ball = 30 Newtons

The distance the ball flew, following the application of the force = 300 meters

Required;

The amount of work done on the ball by Daniel

Solution:

  • Work done = Force applied × Distance moved in the direction of the force

Therefore;

The work done by Daniel on the ball, <em>W</em>, can be calculated as follows;

W = 30 N × 300 m = 9000 J = 9 kJ

  • The work done by Daniel on the ball is 9 kJ

Learn more about work energy and power here:

brainly.com/question/3750360

brainly.com/question/19578325

brainly.com/question/1634438

#SPJ1

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A wooden block is pulled at a constant acceleration of 1.4 m/s2. Find the net applied force on the block if its mass is 0.6 kg.
neonofarm [45]

Answer:

F = 0.84 [N]

Explanation:

In order to solve this problem we must take into account Newton's second law, which tells us that the sum of all forces applied on a body must be equal to the product of the mass of the body by its acceleration.

F = m*a

where:

F = force [N]

a = acceleration = 1.4[m/s^2]

m = mass = 0.6 [kg]

F = 0.6*1.4

F = 0.84 [N]

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A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reach
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Answer:489 Revolutions

Explanation:

Given

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0=96+(-1.5)t

t=64 sec

angular displacement is given by

\theta =\omega_0t+\frac{1}{2}\alpha t^2

\theta =96(64)-\frac{1}{2}(-1.5)(64^2)=3072 degree

For revolutions =\frac{3072}{2\cdot \pi}=488.86 \approx 489 revolution during Slowdown

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