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aliina [53]
3 years ago
6

A 140 g tennis ball travelling 30° east of north at 15 m/s is struck by a tennis racquet, giving it a velocity of 25 m/s, west.

What are the magnitude and direction of the impulse given to the ball? Include a labelled diagram.
Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

The magnitude of the <u>impulse is 1.4 Ns in the West direction.</u>

Explanation:

The impulse equation is given by:

\Delta p=m\Delta v

Where:

  • m is the mass of the ball
  • Δv is the difference in velocity

So we will have:

\Delta p=0.140(25-15)

\Delta p=1.4\: Ns

Therefore, the magnitude of the <u>impulse is 1.4 Ns in the West direction.</u>

I hope it helps you!

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An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
4 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

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Dennis_Churaev [7]

Answer:

Explanation:

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COP = T2 / (T1 - T2)

Where, T1 be the temperature of hot reservoir, T2 be the temperature of cold reservoir.

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jasenka [17]

Answer:

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Answer: Non metal.

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