Answer:
Explanation:
Let the balls collide after time t .
distance covered by falling ball
s₁ = v₀ t + 1/2 g t²
distance covered by rising ball
s₂ = v₀ t - 1/2 g t²
Given ,
s₁ + s₂ = D
D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²
= 2v₀ t
t = D / 2v₀
s₂ = v₀ t - 1/2 g t²
= v₀ x D / 2v₀ - (1/2) x g x D² / 4v₀²
= D / 2 - gD² / 8 v₀²
Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water,
= 5.50 m/s
Her speed after entering the water,
= 1.1 m/s
Time = 1.65 s
Using equation of impulse,

Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
![$= m \times [V_f - V_o] $](https://tex.z-dn.net/?f=%24%3D%20m%20%5Ctimes%20%5BV_f%20-%20V_o%5D%20%24)
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be


= 218 N
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
Answer:

Explanation:
given,
turntable rotate to, θ = 5 rad
time, t = 2.8 s
initial angular speed = 0 rad/s
final angular speed = ?
now, using equation of rotational motion



α = 1.28 rad/s²
now, calculation of angular velocity



hence, the angular velocity at the end is equal to 3.584 rad/s
Answer: A
Hope this help you!!