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ratelena [41]
3 years ago
6

Pluto is different from the outer planets in all of the following ways except which one?

Physics
1 answer:
sergey [27]3 years ago
5 0

Answer:

it is a planet but much smaller

Explanation:

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5. A ball is sitting at rest on the floor. What will happen if a balanced force is applied to the ball? *
Leno4ka [110]

When the balanced force is applied on the ball It will roll away from the force.

<u>Explanation:</u>

  • A ball lies on the floor in rest. If the balanced force is applied to

         the ball, the force will push away.

  • The forces would include gravity and the forces of air particles entering the ball from almost all directions.
  • And the ground is exercising the force and shifting away from the impact.

<u></u>

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3 0
4 years ago
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the
Vladimir [108]

Answer:

Initial velocity = 29.43m/s b) Height it reaches = 44.145 m

Explanation:

Using the first equation of motion we have

v=u+at

here

v is the final velocity

u is the initial velocity

a is the acceleration of the object

t is time

When the ball reaches it's highest point it's velocity will become 0 as it will travel no further

Also the acceleration in our case is acceleration due to gravity (-9.8m/s^{2}) as the ball moves in it's influence alone with '-' indicating downward direction

Thus applying the values we get

0=u-(9.81)m/s^{2}\times 3\\\\u=19.43m/s

b)

By 3rd equation of motion we have

v^{2}-u^{2}=2gs

here s is the distance covered

Applying the value of u that we calculated we get

s=\frac{-u^{2}}{2g}\\\\s=\frac{-29.43^{2}}{-2\times 9.81}\\\\s=44.145m

7 0
3 years ago
A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

4 0
4 years ago
HERES THE ANSWER AND QUESTION
tatyana61 [14]

Answer:

agree

Explanation:

5 0
3 years ago
Read 2 more answers
If the length of a ramp is increased, what will happen to the input force?
Art [367]
The input force will increase too because the more the ramp is increasing the more force is being put on the object going down the ramp... hope this helps ^-^
3 0
3 years ago
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