Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
P = m*v
conservation of momentum suggests
initial momentum equals final momentum
mv-initial = mv-final
(0.0010 kg)(50 m/s) = (0.0010 kg + 0.35 kg)v
thus:
v = (0.0010)(50)/(0.351) = 0.142 m/s
Answer:
v = 1.98*10^8 m/s
Explanation:
Given:
- Rod at rest in S' frame
- makes an angle Q = sin^-1 (3/5) in reference frame S'
- makes an angle of 45 degree in frame S
Find:
What must be the value of v if as measured in S the rod is at a 45 degree)
Solution:
- In reference frame S'
x' component = L*cos(Q)
y' component = L*sin(Q)
- Apply length contraction to convert projected S' frame lengths to S frame:
x component = L*cos(Q) / γ (Length contraction)
y component = L*sin(Q) (No motion)
- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be equal:
L*sin(Q) = L*cos(Q) / γ
Given: γ = c / sqrt(c^2 - v^2)
c / sqrt(c^2 - v^2) = cot(Q)
1 - (v/c)^2 = tan(Q)
v = c*sqrt( 1 - tan^2 (Q))
For the case when Q = sin^-1 (3/5)::
tan(Q) = 3/4
v = c*sqrt( 1 - (3/4)^2)
v = c*sqrt(7) / 4 = 1.98*10^8 m/s
