Answer:
h = 120 m
h = 60 m
when rounded to two significant digits.
Explanation:
height of the higher ball drop point
h = ½gt² = ½(9.8)5.0² = 122.5 m
lower ball drop point
h = ½(9.8)(5.0 - 1.5)² = 60.025
Answer:
0.09 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
From the question, expression for acceleration is given as
F' = ma
Using Pythagoras Theory,
√(F₁²+F₂²) = ma................... Equation 1
Where F₁ = Force of the First person on the boulder, F₂ = Force of the Second person on the boulder, F' = resultant force acting on the boulder, m = mass of the boulder, a = acceleration of the boulder.
make a the subject of the equation
a = √(F₁²+F₂²) /m................ Equation 2
Given: m = 825 kg, F₁ = 64 N, F₂ = 38 N,
Substitute into equation 2
a = [√(64²+38²)]/825
a = {√(5540)}/825
a = 74.43/825
a = 0.09 m/s²
Answer:
The final velocity of the race car is 27.14 m/s
Explanation:
Given;
initial velocity of the race car, u = 18.5 m/s
acceleration of the race car, a = 2.47 m/s²
distance covered by the race car, s = 79.78 m
Apply the following kinematic equation to determine the final velocity of the race car.
v² = u² + 2as
v² = (18.5)² + 2(2.47)(79.78)
v² = 736.363
v = √736.363
v = 27.14 m/s
Therefore, the final velocity of the racecar is 27.14 m/s
Answer:
12.974 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of friction =0.66
a = Acceleration = 
Car's original speed before braking was 12.974 m/s
