Answer:
depends on what type of car it is
Explanation:
Answer:
a) variation of the energy is equal to the work of the friction force
b) W = Em_{f} -Em₀
, c) he conservation of mechanical energy
Explanation:
a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.
Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.
b) the calluses that he would use are to hinder the worker's friction force and energy
W = Em_{f} -Em₀
N d = ½ m v² - m g (y₂-y₁)
y₂-y₁ = 35 -10 = 25m
c) if there is no friction, the physical principle is the conservation of mechanical energy
If there is friction, the principle is that the non-conservative work is equal to the variation of the energy
Answer:
Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:
Answer:
N = 1364 N
Explanation:
given data
accelerate upward = 5.70 m/s²
mass = 88.0 kg
solution
normal force is in upward direction so, weight of the student in downward direction and acceleration is in upward direction so formula is express as
N - mg = ma ...........................1
N = m × (g+a)
put here value
N = 88.0 × (9.8 + 5.70)
N = 1364 N
Answer:
Pascal Law's says that:
If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:
F' = F*(A'/A).
b) Now we can apply this to our particular case:
if the area of one end is 0.01m^2, and the area of the other end is 1m^2
Then we have:
A = 0.01m^2
A' = 1m^2
So, if now we apply a force F in the first end, the force experienced at the other end will be:
F' = F*(1m^2/0.01m^2) = F*100
This means that the force in the other end must be 100 times the force in the first end.
