Answer:
a)
& 
b) 
c) 
Explanation:
Given:
mass of the book, 
combined mass of the student and the skateboard, 
initial velocity of the book, 
angle of projection of the book from the horizontal, 
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

where:
final velcotiy of the student after throwing the book
b)



c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.



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:
1.98 atm
Explanation:
Given that:
Temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28 + 273.15) K = 301.15 K
n = 1
V = 0.500 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K
⇒P (ideal) = 49.45 atm
Using Van der Waal's equation
R = 0.0821 L atm/ K mol
Where, a and b are constants.
For Ar, given that:
So, a = 1.345 atm L² / mol²
b = 0.03219 L / mol
So,


⇒P (real) = 47.47 atm
Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm