Answer:D
Explanation:
Given
mass of A is twice the mass of B half the velocity of B
Suppose
and
be the average force exerted on A and B respectively
and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.
Thus
and option d is correct
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
The average current density in the wire is given by:

where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:

where r is the radius of the wire. In this problem,
, so the cross-sectional area is

and the average current density is

Answer:
The wife have to sit at 0.46 L from the middle point of the seesaw.
Explanation:
We need to make a sketch of the seesaw and the loads acting over it.
And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.
A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.
The maximum momentum with respect to the pivot point (0) will be:

Where L/2 is the half of the distance of the seesaw
Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

Therefore, the coefficient of kinetic friction is approximately 0.0367