Radiation from the sun reaching earth (just outside the atmosphere) has an intensity of 1.39 kw/m2. (a) assuming that earth (and its atmosphere) behaves like a flat disk perpendicular to the sun's rays and that all the incident energy is absorbed, calculate the force on earth due to radiation pressure. (b) for comparison, calculate the force due to the sun's gravitational attraction. assume that the speed of light and earth radius are 2.998 × 108 m/s and 6.37 thousand km respectively.

Explanation:

Given that

Intensity I

Radius of earth,R = 6370 Km

We know that surface area of earth, A

As we know that pressure due to intensity given as

V =Velocity of light

We know that force F

F = P .A

b)Gravitational force F

So F

(a) 5.91 × 10⁸ N; (b) 3.542 × 10²² N

Explanation:

Data:

  I = 1.39 kW/m² = 1.39 × 10³ N·m⁻³s⁻¹

 c = 2.998 × 10⁸ m/s  

 r = 6370 km = 6.37 × 10⁶ m

G  =  6.674 ×10⁻¹¹ N⋅m²kg⁻²

M = 1.989 × 10³⁰ kg  = mass of Sun

m = 5.972 × 10²⁴ kg = mass of Earth

d = 1.496 × 10¹¹ m     = distance from Earth to Sun

Calculations:  

(a) Force exerted by the radiation pressure

(i) Radiation pressure

All the incident radiation is absorbed, so

(ii) Area of Earth's disc

(iii) Radiation force

(b) Gravitational Force

divide the right side by the left side

0.024÷1.2=0.02 for example

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27