Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A = 
= 
Momentum of block B = 
= 
After collision:
Momentum of block A = 
= 
Applying law of conservation of momentum to find momentum of block B after collision 
.
Plugging in the given values and simplifying.
Adding 200 to both sides.
∴ 
Momentum of block B after collision =
Given: Heat Qout means useful work = 2800 J
Heat Qin = 8900 J
Required; Efficiency = ?
Formula: Efficiency = Qout/Qin = x 100%
= 2800 J/8900 J = 0.3146 X 100 %
Efficiency = 31.46%
I believe that you would weigh around 68 or 69 N, or 7 kilograms.
Answer:
D) 15s
Explanation:
let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.
the period of a pendulum is given by:
T = 2π√(L/g)
so on earth:
Te = 2π√(L/g1)
= 6s
on the moon;
Tm = 2π√(L/g2)
since g2 = 1/6 g1 then:
Tm = 2π√(L/(1/6×g1))
= √(6)×2π√(L/(g1))
and 2π√(L/(g1)) = Te = 6s
Tm = (√(6))×6 = 14.7s ≈ 15s
Therefore, the period of the block-spring system on the moon is 15s.
Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position. The star's apparent motion is called stellar parallax. The distance d is measured in parsecs and the parallax angle p is measured in arcseconds.
I hope this helps!
