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sergeinik [125]
3 years ago
15

Can sb try and do this I’m a mother trying to do school again❤️

Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

dont do it to me i wont do it to you :)

Explanation:

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A bat hasa mads of 2kg at the velocity of 45 m/s what is the kinectic energy could he give to a ball
il63 [147K]

Answer:

the  kinetic energy the bat can give to a ball is 2,025 J.

Explanation:

Given;

mass of the bat, m = 2kg

velocity of the bat, v = 45 m/s

The kinetic energy the bat can give to a ball is calculated as;

K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 2 \ \times \ 45^2\\\\K.E = 2,025 \ J

Therefore, the  kinetic energy the bat can give to a ball is 2,025 J.

8 0
3 years ago
Does anyone know this? How far does it travel?
erik [133]
U=5.82,a=2.35,t=3.25

s=u*t+ \frac{1}{2}*a* t^{2}

just substitute.
5 0
3 years ago
A car has a kinetic energy of 1.9 × 103 joules. If the velocity of the car is decreased by half, what is its kinetic energy? 2.2
Harman [31]
If the velocity is decreased by half, then Kinetic energy will reduced to it's 4 times,
= 1.9 * 10³ / 4
=  4.75 * 10²

In short, Your Answer would be Option B

Hope this helps!
8 0
3 years ago
Read 2 more answers
Acceleration is the magnitude of average velocity true or false
Talja [164]

Answer:

False

Explanation:

8 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
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