Answer:
After walking across a carpeted floor in socks, Jim brings his finger near a metal doorknob and receives a shock. What does this demonstrate? Frictional forces require direct contact. Electrical forces can act at a distance.
Answer:
A) Gravitational Force is greater in S.
B) Time taken to fall a given distance in air will be greater for F.
C) Both will take same time to fall in a vacuum.
D) Total force is greater in S.
Explanation:
(a) In this case, the gravitational force of S will be greater, because Newton's Second Law states that - F = ma, or weight =mg. g is constant. And mass of the solid metal is heavier.
(b) In this case, the time it will take for F to fall from a given distance in air will be greater than that of S, since the air resistance is not negligible (as in the case of S).
(c) In this, It will take same time for S and F because in a vacuum, there are no air particles, so there is no air resistance and gravity is the only force acting and so objects fall at the same rate in a vacuum.
(d) The total force will be greater in S than F because Force=ma and S is of heavier mass than F.
Answer:
Here are a few:
1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.
2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.
3) The nebulae, comet, lens flare, and other junk in the background is incorrect.
4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.
5) The planets are incorrectly scaled both to each other and to the sun.
Answer:
it's zero
Explanation:
it is there is your answer
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
