Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.

5.49 x 10^17 m  is the distance between the sun-like star to the earth

Explanation:

Radiation intensity on Earth = 1.0 x 10^-10 W/m^2

Power of radiation of the star = 3.8 x 10^26 W

Recall that the intensity of radiation is given as

=

where

= intensity of radiation

P = power of radiation

A is the area through which the radiation spreads out in all three dimensional direction.

A = = = 3.8 x 10^36 m^2

This area is spread out in the form of a sphere of area

A = = 4 x 3.142 x

3.8 x 10^36 = 12.568

=  (3.8 x 10^36)/12.568 = 3.02 x 10^35

r = = 5.49 x 10^17 m   this is the distance of the star to the Earth

i don't understand. explain it clearly

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft