KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
B. The current increases.
Explanation:
As we know that rate of flow of charge through the conductor is known as electric current
So we have

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased
so we can say that the ratio of charge and time will increase
so here we have

So correct answer will be
B. The current increases.
Answer:
Explanation:
Given
car A had a head start of 
and it starts at x=0 and t=0
Car B has to travel a distance of 
where
is the distance travel by car A in time t
distance travel by car A is

For car B with speed 



Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
In order to be considered a vector, a quantity must include Magnitude (A) and Direction (D).