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4 years ago

Identify the position where the cyclist has maximum kinetic energy.

Physics

2 answers:

Gnom [1K]4 years ago

6 0

Answer:

the correct answer is the third position

finlep [7]4 years ago

3 0

You need the picture. However, since I'm doing PLATO as well, here's the answer:

The third position in which the cyclist is going downhill is the position where the cyclist has the maximum kinetic energy. The other two photos show that the biker has more potential energy.

Thanks, let me know if this helped!

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Why are the parts of an atom that electrons occupy called electron clouds?

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3 years ago

Which factor indicates the amount of charge on the source charge?

Elina [12.6K]

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3 years ago

Read 2 more answers

A car slows down uniformly from a speed of 22 m/s to rest in 4.0 seconds. How far did it travel in that time

krek1111 [17]

Answer:

$\boxed{\sf Distance \ travelled = 44 \ m}$

Given:

Initial speed (u) = 22 m/s

Final speed (v) = 0 m/s (Rest)

Time taken (t) = 4 seconds

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = (\frac{v + u}{2} )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( \frac{0 + 22}{2} ) \times 4 \\ \\ \sf \implies s = \frac{22}{2} \times 4 \\ \\ \sf \implies s = 11 \times 4 \\ \\ \sf \implies s = 44 \: m$

$\therefore$

Distance travelled by car (s) = 44 m

4 0

4 years ago

A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop

vova2212 [387]

Answer:

t=240s

Explanation:

Distance=120m

Acceleration=-5m/s^2

v=0

Let u=x m/s

Using equation v^2-u^2=2as:-

0-x=2(-5)(120)

-x=-1200

x=1200m/s

Using now equation v=u+at:-

0=1200+-5t

5t=1200

t=240s

8 0

4 years ago

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago