Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
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Explanation:
It is given that,
Speed of the jet airplane with respect to air,
If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, 
(A) Let
is the speed of the airliner relative to the ground if the airplane is flying from west to east,

(B) Let
is the speed of the airliner relative to the ground if the airplane is flying from east to west,

Hence, this is the required solution.
Answer:
0.363999909622
Explanation:
F = Force
m = Mass = 15.6 g
C = Drag coefficient
ρ = Density of air = 1.21 kg/m³
A = Surface area = 
v = Terminal velocity = 
s = Displacement = 150 m

Force is given by
F = ma

The drag coefficient is 0.363999909622 (ignoring negative sign)