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stepladder [879]
3 years ago
15

What colors are in a 24 pack of crayola crayons?

Physics
1 answer:
Xelga [282]3 years ago
3 0
Blue puprle white gold organs yellow red gray brown green pink black tan light blue
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The soccer player decelerates from 10 m/s to 0 m/s. Because he decelerates the acceleration will be a negative value. Since he takes 3 seconds to decelerate, divide his original velocity by 3



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When starting a foot race, a 64 kilogram sprinter exerts an average force of 693 newtons backward on the ground for 0.59 seconds
cluponka [151]

The distance traveled by the sprinter in meters is determined as 1.88 m.

<h3>Acceleration of the sprinter</h3>

The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

where;

  • F is the applied force by the sprinter
  • m is mass of the sprinter
  • a is acceleration of the sprinter

a = 693 N / 64 kg

a = 10.83 m/s²

<h3>Distance traveled by the sprinter</h3>

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

  • u is initial velocity = 0

s = ¹/₂at²

where;

  • t is time of motion
  • a is acceleration

s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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1 year ago
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1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

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