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4 years ago

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Physics

2 answers:

V125BC [204]4 years ago

8 0

D. 128.25 because force=mass x acceleration

gtnhenbr [62]4 years ago

8 0

Answer:

D. 128.25 N

Explanation:

The force exerted on an object and the acceleration of the object are related by the following equation (Newton's second law):

$F=ma$

where

F is the force exerted on the object

m is the mass of the object

a is the acceleration

In this problem, we know:

- The mass of the object: m = 13.5 kg

- The acceleration: a = 9.5 m/s^2

Therefore, we can find the force required to give the object this acceleration:

$F=ma=(13.5 kg)(9.5 m/s^2)=128.25 N$

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Answer:

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3 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

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Explanation:

From the exercise we got the ball's equation of position:

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a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

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$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

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$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

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