Answer:
Explanation:
Let 
be the time required to make one revolution.
Let 
be the radius of the circular path.
Let 
be the distance travelled by ball in one revolution.
As we know,the distance travelled in one revolution is the circumference of the circle.
So,
Given,
Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.
Let 
be the speed of the ball.
So,the speed of the ball is 
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
When A Light Ray Hits A Boundary,,,,, it is called refraction.
They differ from each other<span> in wavelength. Wavelength is the distance between </span>one wave<span> crest to the next. </span>Waves<span> in the </span>electromagnetic<span> spectrum vary in size from very long radio </span>waves<span> the size of buildings, to very short gamma-rays smaller </span>than<span> the size of the nucleus of an atom.</span>
Answer:
Draw a free-body diagram to show all forces acting on the object.
