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pshichka [43]
3 years ago
6

In being served, a tennis ball is accelerated from rest to a speed of 42.1 m/s. The average power generated during the serve is

2920 W. Assuming that the acceleration of the ball is constant during the serve, find the force acting on the ball.
Physics
1 answer:
hoa [83]3 years ago
8 0

Answer:

Force F = 69.35 N

Explanation:

given data

Ball Initial speed u = 0

Ball Final speed v = 42.1 m/s

average power generate = 2920 W

solution

Power generate is express as

P= \frac{W}{t}   ..............1

here W is work done and t is time

and work w = F × d

so

P= \frac{Fd}{t}  

and we know speed v = \frac{d}{t}  

so here

Power P = F × v

put here value and we get force

Force F = \frac{2920}{42.1}

Force F = 69.35 N

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According to the theory of a symmetry breaking, at the moment of quark confinement, when the universe was 10-6 seconds old, what
AlekseyPX

At the time of quark confinement, when the universe was 10-6 seconds old, there is found to be one additional proton for every billion antiprotons.

<h3>What is quark confinement?</h3>

Note that  one quark is never found on its own but if particles are said to be smashed together and quarks are found, they are said to be like ends of rubber bands that expands.

Hence, At the time of quark confinement, when the universe was 10-6 seconds old, there is found to be one additional proton for every billion antiprotons.

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7 0
1 year ago
The _____________ variable is observed, measured, and affected by the independent variable.​
Katen [24]
The dependent variable
6 0
3 years ago
Read 2 more answers
An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

5 0
2 years ago
Please solve this question ​
lesantik [10]

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

8 0
2 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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