Answer:
Correct answer: F₂ = 104.5 N
Explanation:
Given:
m = 57 g = 57 · 10⁻³ kg
Δt = 30 ms = 30 · 10 ⁻³ seconds
V₁ = 73.14 m/s service speed
V₂ = 55 m/s returned speed
M = m · V Momentum or Impulse
You forgot to indicate what time the ball contact when returning.
We will assume that the time is the same Δt = 30 ms = 30 10 ⁻³ seconds.
The formula for calculating force is according to Newton's second law is:
F = ΔM / Δt = m · ΔV / Δt
Force during service is:
F₁ = 57 · 10⁻³ · 73.14 / 30 · 10 ⁻³ = 138.97 N
F₁ = 138.97 N
Returned force:
F₂ = 57 · 10⁻³ · 55 / 30 · 10 ⁻³ = 104.5 N
F₂ = 104.5 N
God is with you!!!
Answer:
The induced current is 0.084 A
Explanation:
the area given by the exercise is
A = 200 cm^2 = 200x10^-4 m^2
R = 5 Ω
N = 7 turns
The formula of the emf induced according to Faraday's law is equal to:
ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt
Replacing values:
ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V
the induced current is equal to:
I = ε /R = 0.42/5 = 0.084 A
Answer:
Air resistance have some significant role in projectile motion if the motion lasts for some time.
Explanation:
- Air resistance or air drag seems to be important in daily actvities and games like baseball.
- The trajectory of the projectile with or without air resistance or air drag is totally different.
- When we neglect air drag, the only acting force is gravity against the motion so the maximum height and range are suppose Hmax and R.
- Now, when we consider air drag, it is important to notice that there are two forces against the motion of the ball and along the direction of gravity. It seems that both maximum height and range are lesser Hmax'< Hmax and R'<R.
Due to the shape of the lens , parallel rays will be deviated
