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Sonja [21]
2 years ago
15

The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st

op the car, how far did the car travel before it stopped
Physics
1 answer:
RSB [31]2 years ago
6 0

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

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A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl
sp2606 [1]

Answer:

7.9m/s

Explanation:

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By using g=9.8m/s^2

a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2

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8 0
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A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2
Harman [31]
<span>When the fuel  of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction. 

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a = 86 m/s^2 
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2) 
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Then, the second distance
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Then, we determine the maximum altitude:
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5 0
3 years ago
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