Let u = the speed of the car at the instant when braking begins.
The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.
Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
u = 27.2546 m/s
Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
= 2.7696 m/s
Answer: 2.77 m/s (nearest hundredth)
Answer:
A) 350 N
B) 58.33 N
C) 35 kg
D) 35 kg
Explanation:
If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2
The weight of the object on Earth is given by:
Weight = mass * g = 35 * 10 = 350 N
The weight of the object on the Moon:
Weight = mass * gmoon = 35 * 5/3 = 58.33 N
The mass of the object on Earth is 35 kg
The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.
200n because it's 2×5=10so maybe try solving the problem like that ok does that help
