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Amiraneli [1.4K]
2 years ago
5

For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas

s of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10-3 s), estimate the average force on the ball. Would this force be large enough to lift a 60-kg person?
Physics
1 answer:
inn [45]2 years ago
6 0

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

F= ma

And we know as well that

a= \frac{v}{t}

Replacing the aceleration value in the equation force we have,

F= \frac{mv}{t}

Substituting our values we have,

F= \frac{(0.060)(55)}{4*10^{-3}}

F=825N

The weight of the person is then,

W = mg

W = (60)(9.8) = 558N

<em>We can conclude that force on the ball is large to lift the ball</em>

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A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.
ad-work [718]

Explanation:

Power is defined as the work done per unit time or

P = \dfrac{\Delta W}{\Delta t}

\;\;\;\;= \dfrac{4000\:\text{J}}{5\:\text{s}} = 800\:\text{Watts}

5 0
2 years ago
Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi
Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

\delta = m\lambda (1)

where

\delta =|d_1 -d_2| is the path difference, with

d_1 is the distance of the point from the first source

d_2 is the distance of the point from the second source

m is an integer number

\lambda is the wavelength

In this problem, we have

d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m

So let's use eq.(1) to see if the resulting m is an integer

\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

\delta = (m+\frac{1}{2})\lambda (2)

If we apply the same procedure to eq.(2), we find

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

d_1 = 91.0 m\\d_2 =221.0 m

So the path difference is

\delta =|91.0 m-221.0 m|=130.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

d_1 = 44.0 m\\d_2 =135.0 m

So the path difference is

\delta =|44.0 m-135.0 m|=91.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3

which is an integer: so, this point is a point of destructive interference.

3 0
3 years ago
A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead
zimovet [89]

Answer:

It must be 4 times high.

Explanation:

  • Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
  • This means, that at any time, the following must be true:
  • ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)

⇒      m*g*h = \frac{1}{2} * m*v^{2}

  • Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.

       So, at the bottom of the ramp, all the gravitational potential energy

      must be equal to the kinetic energy of the car (Defining the bottom of

      the ramp as our zero reference for the gravitational potential energy):

       m*g*h_{1}  = \frac{1}{2} * m*v_{1} ^{2}  (1)

  • Now, let's do v₂ = 2* v₁
  • Replacing in (1) we get:

        m*g*h_{2}  = \frac{1}{2} * m*(2*v_{1}) ^{2} (2)

  • Dividing (2) by (1), and rearranging terms, we get:
  • h₂ = 4* h₁
8 0
3 years ago
You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __
SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy {\rm GPE} of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

5 0
1 year ago
What distance does a police car travel if it is going 3.0 m/s for 20 seconds
Vladimir79 [104]

Answer:

60 meters

Explanation:

If you are going 3 meters in a second, and you are traveling for 20 seconds, you have to multiply

3meters/second*20seconds

cross out the seconds and you have

3 meters*20

60 meters

6 0
2 years ago
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