<em>friction transforms KE into thermal energy (a)</em>
That's why, if it goes on long enough, the moving object actually gets warm.
Answer:
Explanation:
All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units
Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0
The forces ONLY acting on the person would be the force of gravity pulling them down, and the normal force that the elevator is reciprocating from the person standing on it.
Force = mass x acceleration. You have 100 kg and you are accelerating at 0 m/s. The force is 0. Which makes sense because the force of gravity and the net force completely cancel each other out.
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
Explanation:
First of all let's define the specific molar heat capacity.
(1)
Where:
Q is the released heat by the system
n is the number of moles
ΔT is the difference of temperature of the system
Now, we can find n with the molar mass (M) the mass of the compound (m).
Using (1) we have:
I hope it helps!
<h3><u>Answer;</u></h3>
electric potential
<h3><u>Explanation;</u></h3>
Electric potential is the electric potential energy per unit charge.
Mathematically; V =PE/q
Where; PE is the electric potential energy, V is the electric potential and q is the charge.
Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.
