The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
<u>In modern physics</u>, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T
as:
where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.
then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:
As
Then
then
- The factor will the total energy radiated per second per square meter increase = 256
Your answer is 3 ( 1 calcium atom and 2 bromine atoms)
Answer:
(A) 60 J
Explanation:
At state 1
KE₁=100 J
At state 2
KE₂ = 0
U₂=80 J
Given that surface is rough so friction force will act in opposite to the direction of motion
Lets take work done by friction = Wfr
From work power energy
Work done by all forces = Change in kinetic energy
Wfr + U₂=ΔKE
Wfr+80 = 100
Wfr= 20 J
Now when book slides from top position then
Wfr+ U = KEf - KEi
-20 + 80 = KEf-0
KEf= 60 J
(A) 60 J
