Answer:
The applied force is greater than the frictional force.
Explanation:
the chair moves at <u>a constant speed</u><u> </u><u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>not</u><u> </u><u>A</u><u> </u><u>or</u><u> </u><u>C</u><u>.</u>
if there is no friction then the chair <u>would accelerate and it would not be at a constant speed</u><u>.</u>
hence, the only possible answer is B.
Answer: 55.52 *10^-6 C= 55.52 μC
Explanation: In order to solve this question we have to take into account the following expressions:
potential energy stired in a capacitor is given by:
U=Q^2/(2*C) where Q and C are the charge and capacitance of the capacitor.
then we have:
Q^2= 2*C*U=
C=εo*A/d where A and d are the area and separation of the parallel plates capacitor
Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=
=55.52 *10^-6C
Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.
