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Makovka662 [10]
3 years ago
8

A force pair is produced when a tennis racket strikes a tennis ball. Which of the following best explains why the tennis ball do

es not have zero net force acting on it?
-The force exerted on the ball is greater than that exerted on the racket.
-Each half of the force pair acts on a different object.
-The two forces act in the same direction.
-The forces act perpendicular to each other.
Physics
1 answer:
Serga [27]3 years ago
4 0

Answer:

Each half of the force pair acts on a different object.

Explanation:

When a tennis racket strikes a tennis ball a pair force is produced. when the racket strikes the ball the racket exerts an action force on the tennis ball, according to Newton's third law for every action there is an equal and opposite reaction force, as a reaction the ball exert an equal and opposite force on the racket. These forces are often called pair forces.

As the forces acts on different bodies (Action force act on ball and reaction force act on racket) so the net force tennis ball is never zero.

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A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/
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Which one of the following statements concerning the momentum of a system when the net force acting on the system is equal to ze
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D

Explanation:

According to newton's 2nd law rate of change of momentum is directly proportional to the force applied on the body. Since, net Force is zero this means momentum did not change or momentum of the body remained constant.

Hence, the system have constant value of momentum. Therefore, option D is correct.

5 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
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Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
2 years ago
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