Answer:
Explanation:
Change in length of spring = 2.13 m
Component of weight acting on spring = mg sinθ
so
mg sinθ = k x where k is spring constant and x is total stretch due to force on the spring.
Here x = 2.13
mg sin17 = k x 2.13
31 x 9.8 sin17 = k x 2.13
k = 41.7 N/m
b ) In case surface had friction , spring would have stretched by less distance .
It is so because , the work done by gravity in stretching down is stored as potential energy in spring . In case of dissipative force like friction , it also takes up some energy in the form of heat etc so spring stretches less.
So we can know what is in space maybe weird or interesting stuff
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q = 
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q = 
q = 3.6 10⁵ C
The answer is increased. A power factor of one or "unity power factor" is the aim of any electric utility business from the time when if the power factor is less than one, they have to give more current to the user for an assumed amount of power use. In so doing, they gain more line harms.
