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german
3 years ago
15

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

r starts from rest and accelerates at 3.9m/s/s, how far away do the cars meet up again?
Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

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Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

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Angle = 30.0°

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Coefficient of kinetic friction = 0.50

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U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

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Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

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10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.
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What is the force of a object that has a mass of 7 kg and an acceleration of 6 m/s/s
UkoKoshka [18]

Answer:

<h2>42 N</h2>

Explanation:

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This is shown below:

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