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Black_prince [1.1K]
2 years ago
12

You are pushing a heavy box across a rough floor. when you are initially pushing the box and it is accelerating, (a) you exert a

force on the box, but the box does not exert a force on you. (b) the box is so heavy it exerts a force on you, but you do not exert a force on the box. (c) the force you exert on the box is greater than the force of the box pushing back on you. (d) the force you exert on the box is equal to the force of the box pushing back on you. (e) the force that the box exerts on you is greater than the force you exert on the box.
Physics
2 answers:
kotegsom [21]2 years ago
4 0

Answer:

(c) the force you exert on the box is greater than the force of the box pushing back on you.

Explanation:

When a force is exerted on an object such that it starts accelerating without any change in the state of rest of the source of the force that always means that the reaction force of the object on the source is less than the action force.

This condition is in accordance with Newton's second law and Newton's third law.

kodGreya [7K]2 years ago
3 0

The answer is C. If the box is accelerating, that means that the amount of force you are exerting is greater than the force of the box.

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0.0257259766982 m

Explanation:

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h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

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A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
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Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

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No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

where

i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

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e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

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