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PolarNik [594]
3 years ago
7

Help me please! What is the velocity of a 0.1 kg baseball whose momentum is 100?

Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

1000m/s

Explanation:

velocity(v)=momentum(p)/mass

=100/.1

=1000m/s

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A change in temperature causes a change in density. true or false?
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True a change in temperature can cause a change in density.

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When a gun is fired at the shooting range, the gun recoils (moves backward). Explain this using the law of conservation of momen
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A toy cannon tosses a rubber ball straight upward. A motion sensor measures the speed of the ball as it leaves the cannon. Using
mezya [45]

Answer:

With this information is not possible to calculate the mass.

Explanation:

This is a characteristic problem of energy conservation, where kinetic energy becomes potential energy. For this particular problem, we have the initial speed as input data. The moment the ball comes out of the cannon we have the maximum kinetic energy, as the ball goes up the ball will gain more potential energy as the ball loses kinetic energy, until the moment the ball reaches the maximum height. At the maximum height point, the ball will have its maximum potential energy while its kinetic energy is zero. In other words, all the kinetic energy that was, in the beginning, was transformed into potential energy.

E_{p}=E_{k}\\m*g*h=0.5*m*v^2

In the above equation the masses are canceled and we can determine the maximum height, by means of the initial speed.

h=\frac{0.5*v^2}{g} [m]

But the mass cannot be determined, since it would be necessary to know the value of the energy, in order to determine the value of the mass.

3 0
3 years ago
A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find
Harlamova29_29 [7]

Explanation:

Given,

  • m = 100 kg
  • g = 10 N/kg¹
  • h = 60 m
  • t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

  • a) Work done by the pump

We know that,

\rm \: Work  \: done = Force * Distance  \: moved

  • f = 100 kg * 10N/kg
  • d = 60 m

\rm \: Work\; Done =(100 \: kg \times  \cfrac{10N}{kg} ) \times 60 \: m

\rm \: Work\; Done =1000 \times 60 \: joule

\boxed{\rm \: Work\; Done =60000 \: joule}

[The unit'll be joule since N×M = J]

  • b) Potential energy stored in the water

\rm \: P.E = m \cdot g \cdot  h

  • m = 100 kg
  • g = 10N/kg
  • h = 60

\rm \: P.E =100 \:kg \:  \times  \cfrac{10 \: N}{kg}  \times 60

\boxed{\rm \: P.E =60000  \: joule}

  • same condition here as well, N×M = J
  • c) Power of the Pump

\rm \: P = W/T

  • where P = Power; W = Work done & T = Time taken
  • As we got the value of work done on question (a),& ATQ time taken is 20 S.

\rm \: P =  \cfrac{60000 \: joule}{20 \: seconds}  =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}

  • d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

6 0
2 years ago
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