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jok3333 [9.3K]
3 years ago
9

An electromagnetic wave of intensity 150 W/m2 is incident normally on a rectangular black card with sides of 25 cm and 30 cm tha

t absorbs all the radiation. Find the force exerted on the card by the radiation.
Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

3.75 × 10⁻⁸ N

Explanation:

Given:

Intensity of the electromagnetic wave, I = 150 W/m²

Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)

therefore,

the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²

Now,

force exerted on the card by the radiation, F = \frac{IA}{C}

here,

C is the speed of the light = 3 × 10⁸ m/s

on substituting the respective values, we get

F = \frac{150\times0.075}{3\times10^8}

or

F = 3.75 × 10⁻⁸ N

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An imbalance of electrons causes an object to be charged. A charged substance is one that has a net charge that is not zero. If the number of protons and that of electrons is equal in an object, the object is not neutral, but it has zero charge. Loss or gain of electrons determine the charge of an object, when an object loses electrons it becomes positively charged, but when it gains electrons it becomes negatively charged.
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2. Kevin works as a janitor, and he is pushing a fully-
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The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

  • <em>weight of the load, w = 557 N</em>
  • <em>force applied , F = 410 N</em>
  • <em>angle of force, =  15°</em>
  • <em>coefficient of kinetic friction  = 0.46</em>
  • <em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

Fcos\theta - F_k = ma

where;

  • <em>m is the mass of the recycling bin</em>
  • <em />F_k<em> is the frictional force </em>

W = mg

557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg

The net horizontal force on the recycling bin is calculated as;

Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n  = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2

The time taken for him to move the bin 6.5 m is calculated as follows;

s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2}  \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

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Answer:

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