Objects 1 and 2 attract each other with an electrostatic force of 72.0 units. if the charge of object 1 is halved, then the electrostatic force is 36.0 units
Let's understand the answer
The electrostatic force is the force that exists between two charged particles which is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the distance between the charges
In this problem, the charge of object 2 is doubled, so
So, the electrostatic force will be halved.
Since as the initial electrostatic force is
F = 72.0 units
the new force will be
72/2 = 36 units
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To solve this exercise it is necessary to apply the concepts related to the Snells law.
The law defines that,
Incident index
Refracted index
= Incident angle
Refracted angle
Our values are given by
Refractory angle generated when light passes through the fiber.
Replacing we have,
Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,
Therefore the critical angle for the light ray to remain insider the fiber is 28.6°
It’s important because if you were let’s say training to see how far you go in 10sec you need to record your data to see how much you are improving in that skill or something
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.
