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Serjik [45]
4 years ago
12

8) A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the

Physics
1 answer:
Vlada [557]4 years ago
4 0

8)Acceleration of the car is 11.17 m/s².

9) Acceleration of skater is 0.57 m/s².

10) The deceleration is 1.8 m/s².

11) The average acceleration of the sprinter is 2 m/s².

12) The velocity of the stroller after 4.75 minutes is 171 m/s.

13) The skier will be moving as 11 m/s speed.

14) 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

Answer:

Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

Acceleration = (46.1-18.5)/2.47 = 11.17 m/s².

Acceleration of the car is 11.17 m/s².

9)Similarly, the acceleration of the skater with initial velocity zero to final velocity as 5.7 m/s in 10 seconds is

Acceleration = (5.7-0)/10=0.57 m/s².

So acceleration of skater is 0.57 m/s².

10) In this case, the shuttle bus is stopping so its speed is decreasing from high value to low value. This kind of acceleration related to the decrease in velocity is termed as deceleration as the value of acceleration will be coming as negative.

Acceleration = (0-9)/5=-1.8 m/s².

So the deceleration is 1.8 m/s².

11) Average acceleration of sprinter = (Final velocity-Initial velocity)/Time

Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

13) Similarly, in this case, u = 0, a = 2.2 m/s² and t = 5 s

Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

So 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

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The minimum stopping distance when the car is moving at

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Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

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4 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
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Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

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b )

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c ) If  the average frictional force that acts on the sliding bear be F

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then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

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The mechanical advantage of the crowbar is 15.
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