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Sindrei [870]
2 years ago
8

Can somebody please help me? PLEASE

Physics
1 answer:
ale4655 [162]2 years ago
6 0

Answer:

The 1st Answer

Explanation:

Because kinetic energy is the energy which a body possesses by virtue of being in motion. So if the velocity of the object (cannonball in this case) decreases than so would the kinetic energy

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Can the force of gravity cause any deformation to an object?
PSYCHO15rus [73]

Answer:

yes it could deform a shape or an object

Explanation:

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3 years ago
As an object rolls downhill, some of the energy is
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When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)


7 0
3 years ago
Read 2 more answers
1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin
Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0
3 years ago
CAN SOMEONE PLS HELP ME!!!
scZoUnD [109]

Answer:

c

Explanation:

the question answered itself.

4 0
2 years ago
Read 2 more answers
A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a fla
fomenos

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}

We know that range of a projectile is given by

R=\frac{v_{o}^{2}sin(2\theta )}{g}

it is given that R=130 m applying values in the above equation we get

R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s

Thus the force is obtained as

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}

Thus force equals F=338.66N

4 0
3 years ago
Read 2 more answers
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