I think everything looks good but 2 might not be right, it might be B
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
Answer:
Time period for Simple pendulum, 
Explanation:
The Simple Pendulum
Consider a small bob of mass 
is tied to extensible string of length 
that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let 
is the angle made by string with vertical during oscillation.
Vertical component of the force on bob, 
Negative sign shows that its opposing the motion of bob.
Taking as very small angle then, 
Let 
is the displacement made by bob from its mean position ,
then, 
so, 
........(1)
Since, pendulum is in hormonic motion,
as we know, 
where 
is the constant and 
.........(2)
From equation (1) and (2)
Since, 
