Answer:
(a) the force is 8.876 N
(b) the magnitude of each charge is 4.085 μC
Explanation:
Part (a)
Given;
coulomb's constant, K = 8.99 x 10⁹ N.m²/C²
distance between two charges, r = 10 cm = 0.1 m
force between the two charges, F = 15 N
when the distance between the charges changes to 13 cm (0.13 m)
force between the two charges, F = ?
Apply Coulomb's law;

Part (b)
the magnitude of each charge, if they have equal magnitude

where;
F is the force between the charges
K is Coulomb's constant
Q is the charge
r is the distance between the charges

Answer : The mass of ice melted can be, 3.98 grams.
Explanation :
First we have to calculate the moles of ice.

where,
Q = energy absorbed = 27.2 kJ
= enthalpy of fusion of ice = 6.01 kJ/mol
n = moles = ?
Now put all the given values in the above expression, we get:


Now we have to calculate the mass of ice.

Molar mass of ice = 18.02 g/mol

Thus, the mass of ice melted can be, 3.98 grams.
Explanation :
It is given that,
BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :
We know that, 1 kilocalorie = 4184 joules
So, 

J/sec is nothing but watts.
So, 
and 
So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:
