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:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
That is the answer to the question
I hope this helps you.
Thank you for your question
Answer:
65.2 %
Explanation:
Let Q1 = Heat absorbed by the system
Q2 = Heat released by the system
e= (1 - (Q2/Q1)) x 100
e= (1 - (750/2150)) x 100
e= (1 - 0.348) x 100
e= 0.652 x 100
e= 65.2 %
