Answer:
7. Net constant force down the ramp
Explanation:
After the car is released and starts moving up the ramp, the only force that is applied on the car is weight because of the gravity, we were told that the friction force is neglected. the force because of the weight is given by:
where θ is the angle of the ramp.
as you can see those values won't change, so the force remains constant down the ramp.
The magnitude of a vector is the square root of its components squared. When you square a negative number, you will always get a positive number. Its like speed of your car. It will always be positive (or 0) whether you are moving forward or you are in reverse gear or stationary.
D is definitely the correct choice here.
I think a case could also be made for choice-B, but that would be a tough, complex operation.
Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
