Answer:
6.26 m/s
Explanation:
Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)
PE = KE
mgh = 1/2 mv^2 divide both sides of the equation by 'm'
gh = 1/2 v^2 multiply both sides by 2
2 gh = v^2 take sqrt of both sides
v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s
Answer:
The maximum temperature rise = 0.047 °C
Explanation:
Potential Energy, P = mgh
Energy transfered, Q=mcΔT
Potential energy = Energy transfered
mgh = mcΔT
gh = cΔT
ΔT = gh/c
ΔT = (9.81 * 20) / 4186
ΔT = 0.047 °C
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N
In nomine patris, et filii, et spiritus sancti.
