The correct answer to the question is C i.e C represents the friction from air resistance.
EXPLANATION:
Before coming into any conclusion, first we have to understand friction.
The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.
The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.
In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically. There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.
As the plane is moving along the direction of A, the frictional force must act along the direction of C.
Answer:
7.24 ohm
Explanation:
Let R1 and R2 are resistance of two resistors.
Emf=E=20 V
Current,I=2 A
Current,I'=10 A
We have to find the magnitude of the greater of the two resistances.
In series


By using the formula

...(1)
In parallel







Substitute the value




By using quadratic formula




Substitute the value


Hence, the magnitude of the greater of the two resistance=7.24 ohm
The same 500N, is the Newton’s Third Law.
Answer: The molar heat capacity of aluminum is 
Explanation:
As we know that,
.................(1)
where,
q = heat absorbed or released
= mass of water = 130.0 g
= mass of aluminiunm = 23.5 g
= final temperature
=
= temperature of water =
= temperature of aluminium =
= specific heat of water= 
= specific heat of aluminium= ?
Now put all the given values in equation (1), we get
Molar mass of Aluminium = 27 g/mol
Thus molar heat capacity =