Answer:
0.47 N
Explanation:
Here we have a ball in motion along a circular track.
For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.
This force is called centripetal force, and its magnitude is given by:

where
m is the mass of the object
is the angular velocity
r is the radius of the circle
For the ball in this problem we have:
m = 40 g = 0.04 kg is the mass of the ball
is the angular velocity
r = 30 cm = 0.30 m is the radius of the circle
Substituting, we find the force:

Answer:
v=115 m/s
or
v=414 km/h
Explanation:
Given data

To find
Terminal velocity (in meters per second and kilometers per hour)
Solution
At terminal speed the weight equal the drag force

For speed in km/h(kilometers per hour)
To convert m/s to km/h you need to multiply the speed value by 3.6
We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as


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In physics, gravity is the natural force that causes things to fall toward the earth. The noun gravity can also mean seriousness or solemnity. Someone who conducts themselves with an air of gravity is someone who takes what they are doing seriously.
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as

substituting values


The time elapsed in the spacecraft’s frame is mathematically evaluated as

substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)

=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame