Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s
Answer:
4.17 m/s
Explanation:
To solve this problem, let's start by analyzing the vertical motion of the pea.
The initial vertical velocity of the pea is

Now we can solve the problem by applying the suvat equation:

where
is the vertical velocity when the pea hits the ceiling
is the acceleration of gravity
s = 1.90 is the distance from the ceiling
Solving for
,

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

Therefore, the speed of the pea when it hits the ceiling is

Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:

Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction

P=25x10^6 andF=750.So plug in everything to solve for A. which is 3x10^-5m^2 OR 0.3mm^2