Answer:
1) k = 52 N/m
2) E = 1.0 J
3) ω = 8.1 rad/s
4) v = 1.4 m/s
Though asked for a velocity, we can only supply magnitude (speed) because we don't have enough information to determine direction.
If it happens to be the first time it is at y = - 10 cm after release, the velocity is upward.
Explanation:
Assuming the initial setup is after all transients are eliminated.
kx = mg
k = mg/x = 0.8(9.8) / 0.15
k = 52.26666.... ≈ 52 N/m
E = ½kA² = ½(52)(0.20²) = 1.045333... ≈ 1.0 J
ω = √(k/m) = √(52 / 0.8) = 8.0829... ≈ 8.1 rad/s
½mv² = ½kA² - ½kx²
v = √(k(A² - x²)/m) = √(52(0.20² - 0.10²)/0.8) = 1.39999... ≈ 1.4 m/s
Answer:
0.237 (23.7 %)
Explanation:
The thermal efficiency of an engine is given by:

where
W is the useful work output of the engine
is the heat in input
Here we have:

and the work done is the total heat in input minus the heat exhausted:

So, the efficiency is
(23.7 %)
Work done by force is given by formula

now here work done against friction is given so force of friction here is

It moved by three corridors with each measures

so total distance will be

now from above formula of work done we will say


so above is the work done to move three corridors
A : A nebular cloud of dust and gas is the correct answer.
The answer is 13.5 because 27÷3.0=13.5