Answer:
At an angle of 
Explanation:
Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.
The resultant speed will be given by

Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
Answer:

Decrease
Explanation:
I = Current = 3.7 A
e = Charge of electron = 
n = Conduction electron density in copper = 
= Drift velocity of electrons
r = Radius = 1.23 mm
Current is given by

The drift speed of the electrons is 

From the equation we can see the following

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.