The kinetic energy of the block when it reaches the bottom is 39.2 J.
<h3>
Kinetic energy of the block at the bottom</h3>
Apply the principle of conservation of energy.
K.E(bottom) = P.E(top)
P.E(top) = mgh
where;
- m is mass of the block
- g is acceleration due to gravity
- h is the vertical height of fall
P.E(top) = 5 x 9.8 x 0.8
P.E(top) = K.E(bottom) = 39.2 J
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Answer: -7J
Explanation:
To determine the change in potential energy, use the equation ΔPE=mgΔh
Δ
PE=mg
Δ
h
, where m is the mass, g= 9.8 m/s2
g= 9.8 m/s
2
, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J
PE=4.0 kg
×
9.8 m/s
2
×
-2.0 m=-78 J
. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.
Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
Answer:
imma be honest I dint really know
