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:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Answer:
1st, 2nd, and 4th
Explanation:
1st conserves gasoline/petroleum
2nd conserves electricity
4th conserves paper
Answer:
Fossil Combustion Reactions
Explanation:
It's more efficient (I'll edit later)
Answer:
% reduction in area = 54.26 %
percentage elongation = 43.16 %
Explanation:
a) percentage reduction in area 


% reduction in area =
b)percentage elongation 