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3 years ago

an object with a mass of 7.5 kg accelerates 8.3 m/s2 when an unknown force is applied to it. what is the amount of the force?

1 answer:

5 0

F=MA
F is force
M is mass
A is acceleration
F=MA
F=(7.5 kg)(8.3 m/s^2)
F=62.25 N (newton- SI unit for force)
Hope this helps

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What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?

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Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{135}{25} = \frac{27}{5} \\$

We have the final answer as

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3 years ago

A suitcase has a width of 55 cm. Which unit conversation fraction should you use to find the width in inches?

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Answer:

2,54 cm are equal to 1 inch

Explanation:

Doing the conversion:

$55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]$

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3 years ago

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3 years ago

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You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

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The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

$v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt$

The time of motion of the object from the given height is calculated as;

$h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }$

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

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2 years ago

A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th

lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

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3 years ago

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