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3 years ago

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

Physics

2 answers:

posledela3 years ago

5 0

Answer:

<h2>100 newtons</h2>

Explanation:

r-ruslan [8.4K]3 years ago

4 0

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

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