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Readme [11.4K]
3 years ago
5

If a bat with a mass of 5 kg and acceleration of 2 m/s2 hits a ball whose mass is 0.5 kg in the forward direction, what is the r

eaction force of the ball on the bat?
10 N, forward
0 N, forward
0.5 N, upward
10 N, backward
Physics
2 answers:
ivanzaharov [21]3 years ago
8 0

Answer:

10 N , forward

Explanation:

F=m×a

F= 5×2

F=10 N

lina2011 [118]3 years ago
6 0

Answer:

10 N, backward

Explanation:

Force exerted on the ball by the bat can be calculated according to newtons second law,

Force = mass × acceleration

Given mass of the ball = 5kg

Acceleration of the ball = 2m/s²

Force exerted on the ball= 5×2

Force exerted on the ball by the bat = 10N

If the force of 10N is applied on the ball by the bat in the forward direction, the reaction force of the ball on the bat will also be 10N but in the opposite direction (backward) to the direction of the bat according to newton's third law which states that actions and reaction are equal and opposite.

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Which of the following statements is correct?
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Answer: c. Generally, metals are ductile.

Explanation:

From the options given in the question, the correct statement is that"Generally, metals are ductile.

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3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
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Answer:

0.28802

2.57162 W

14.28 W

53.55 W

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Explanation:

R = 280Ω

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V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

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The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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