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Kaylis [27]
3 years ago
11

A 1,500-kg car accelerates along a flat track to a speed of 20 m/s. how much does it's gravitational potential energy increases

over the length of the track?

Physics
1 answer:
lesya [120]3 years ago
3 0
Look at bitesizes physics section, they have all the information you need to complete this question.
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6 0
3 years ago
Read 2 more answers
2. Using examples to describe each term, differentiate
melisa1 [442]

Answer: find the answer in the explanation

Explanation:

Average velocity is the average value of magnitude of initial velocity and final velocity.

If U = initial velocity and V = final velocity, then average velocity can be expressed as

Average velocity = ( U + V )/2

A vehicle who takes 60 minutes to cover 30 miles north and then 30 miles south and end up at the same place, has an average speed of 60 miles divided by 60 minutes, or 1 mile per minute.

Instantaneous velocity is the distance covered in a specific direction per time taken. Instantaneous velocity can be expressed as

Velocity = displacement/ time.

Uniform velocity occurs when we have a constant velocity. That is, when velocity does not change. When a vehicle travels in equal distances in equal intervals of time.

7 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
What is t=3seconds in scrience
alekssr [168]
Well....t usually stands for time and the unit seconds proves that
8 0
3 years ago
A cheetah is resting when a hare traveling in a straight path passes at it's top speed of 25m/s. It takes the cheetah 1.5 second
timofeeve [1]

Answer:

a) 2.3 m/s^2

b) 33.3 s

c) 803 m

d) 0 m/s

Explanation:

We need to apply the accelerated motion formulas.

The acceleration is given by:

a=\frac{v_f-v_o}{t}\\\\a=\frac{45m/s-0m/s}{20s}\\\\a=2.3m/s^2

we need to calculate the time the cheetah takes to stop when it starts to decelerate:

v=vo+a*t\\\\t=\frac{v-v_o}{a}\\\\t=\frac{0-45m/s}{-4m/s^2}=11.3s\\\\t_t=20s+2s+11.3s=33.3s

We need to calculate the displacement for all of the stages:

stage 1:

x_1=\frac{1}{2}*a*t^2\\\\x_1=\frac{1}{2}*2.3m/s^2*(20)^2\\\\x_1=460m

stage 2:

x_2=v*t\\x_2=45m/s*2s=90m

stage 3:

x_3=v_o*t+\frac{1}{2}*a*t^2\\\\x_3=45m/s*(11.3s)-\frac{1}{2}*4m/s^2*(11.3)^2\\\\x_3=253m

The total displacement is given by:

d=x_1+x_2+x_3\\d=460m+90m+253m=803m

the final velocity of the cheetah is zero because the cheetah decelerates until stop.

4 0
4 years ago
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