-- more mass involved
-- less distance between the two objects
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: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.
Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo
E* 4*π*r^2= Q inside/εo
E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C
D. it is thickest in the middle
Answer:
As b ∝ (L/r²) and
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
hence,
the Sun appear brighter in the sky
Explanation:
The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).
thus, mathematically,
b ∝ (L/r²)
now,
given
L for sirius is 23 times more than the sun i.e 23L
now,
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
thus,
using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.
hence, the sun appears brighter