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Ne4ueva [31]
2 years ago
7

I need help plz!!! Please help...

Physics
1 answer:
RSB [31]2 years ago
4 0

Answer: 60

Explanation:

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc
alexdok [17]

Answer:

20.25 m

Explanation:

  • <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

         <em><u>ac = v²/r</u></em>

<em>         </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m</u></em>

Therefore;

r = v²/ac

  = 27²/36

  = 20.25 m

Hence the radius is 20.25 meters

5 0
3 years ago
Read 2 more answers
What evidence do you that suggest water waves are transverse wave​
svp [43]

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

5 0
3 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
There are ______ main fitness zones.
vlabodo [156]

Answer:

option D

Explanation:

five

hope it helps you and Mark me down as brainlist

3 0
3 years ago
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