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siniylev [52]
3 years ago
14

A pitcher exerts 100.0 N of force on a ball with a velocity of 45 m/s. What is the pitcher's power?

Physics
1 answer:
yarga [219]3 years ago
8 0
We use the relation:

Power = force x velocity
Power = 100 x 45
Power = 4,500 Watts
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HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!
soldi70 [24.7K]
Limestone, Sandstone, and Shale would be the answer.

4 0
3 years ago
Phyics !!! Cannon ball question
natima [27]

Answer:

Zero; no force is required to keep it going

Explanation:

Since the cannon ball is fired into frictionless space, there will be nothing to stop it, so it will keep going and going.

3 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
Please someone do this <br>please ​
monitta

The various contributions involved till the chapati is made is given below.

<h3>What is food?</h3>

The substance that we intake for the body to charge up by giving nutrients is called the food.

Wheat is a staple food. We make chapati from flour obtained from the wheat grains.

The various contributions involved till the chapati is made is given below.

                 Take required amount of atta in a container

                                                     ↓

                    Add water accordingly to form a dough

                                                     ↓

                 Apply oil to make dough smooth for long time

                                                    ↓

        Take small dough, make it a ball shaped and apply dry flour

                                                    ↓

               Roll it using rolling pin on the chapati maker plate

                                                    ↓

      After making it circular or any shape you want, place it on hot tawa

                                                    ↓

                               Bake it on both the sides

                                                   ↓

                                      Chapati is ready

Thus, the flow chart is made.

Learn more about food.

brainly.com/question/16327379

#SPJ1

6 0
1 year ago
A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull
sdas [7]

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
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