You drop a 5 kg ball from a height of 2 m. Justbefore it reaches the ground, how much kineticenergy does it have?

12 answers
Question:

You drop a 5 kg ball from a height of 2 m. Just before it reaches the ground, how much kinetic
energy does it have?

Answers

what is this.. .. .. . ur question is not clear

A man drops a 10 kg rock from the top of a

S rock from the top of a 5 m ladder. What is its speed just before it hits

What is its kinetic energy when it reaches the ground. 18

Explanation:

Kinetic  energy = 98 N

Explanation:

Given:

Mass = 5 kg

Height = 2 m

Find:

Kinetic  energy

Computation:

K(e) = P(e)

So,

P(e) = mgh

So,

K(e) = (5)(9.8)(2)

Kinetic  energy = 98 N

6

Explanation:

0.5(3 × 2^2)

Explanation:

mass(m)=2 kg

velocity(v)=5m/s

Now,

Kinetic energy(K.E.)=1/2mvsq

=1/2×2×5sq

=25

where,sq=square(2)

Given:

m = 0.32 kg

v = 11.5 m/s

To find:

1) Kinetic energy = ?

2) Work needed to stop the ball = ?

Formula used:

1) Kinetic energy = [tex]\frac{1}{2} m v^{2}[/tex]

2) Work = kinetic energy

Solution:

1)

Kinetic energy is given by,

Kinetic energy = [tex]\frac{1}{2} m v^{2}[/tex]

K.E. = [tex]\frac{1}{2} 0.32 \times 11.5 \times 11.5[/tex]

K.E. = 21.16 Joule

Thus, option (d) is the correct answer.

2)

Work needed to stop the ball is same as the kinetic energy because energy is the capacity to do work. Since, work is opposing the movement of the ball. Thus, Work = - Kinetic energy

Work = -21.16 Joule

Here's link to the

tinylnk.cf/rW5p

KE = 9720 J

KE = 0.00972 MJ

so it would be 97.2

See explanations below

Explanation:

1) KE = 1/2mv²

mass m = 0.142kg

v = 40m/s

KE = 1/2 * 0.142 * 40^2

KE = 1/2 * 0.142 * 1600

KE = 800*0.142

KE = 113.6Joules

2) KE = 1/mv^2

KE = 196

mass m = ?

velocity v = 52m/s

Substitute

196 = 1/2*m*52²

196 = 1352m

m = 196/1352

m = 0.145kg

Hence the mass of the baseball is 0.145kg

3) KE = 1/2mV^2

288 = 1/2 * 4v²

288 = 2v²\

v² = 288/2

v² = 144

v = 12m/s

The velocity of the shotput is 12m/s

If you drop a 6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and after the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45. 

the correct answer is b

i'm on my way to history not a mistory.

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