Velocity<span> is a</span>vector<span> quantity; it is direction-aware.</span>
An applied force<span> is a </span>force<span> that is </span>applied<span> to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
</span>Normal force<span> is the component, perpendicular to the surface (surface being a plane) of contact.
</span><span>The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.</span>
Answer:
Friction always acts opposite to the motion.
iIn this case the mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing the rotational motion
Therefore the answer is False
The options are;
a. V2 equals 2V1.
b. V2 equals (V1)/2.
c. V2 equals V1.
d. V2 equals (V1)/4.
e. V2 equals 4V1.
Answer:
Option A: V2 equals 2V1
Explanation:
Since the flow is steady, then we can say;
mass flow rate at input = mass flow rate at output.
Formula for mass flow rate is;
m' = ρVA
Thus;
At input;
m'1 = ρ1•V1•A1
At output;
m'2 = ρ2•V2•A2
So, m'1 = m'2
Now, we are told that the density of the fluid decreases to half its initial value.
Thus; ρ2 = (ρ1)/2
Since m'1 = m'2, then;
ρ1•V1•A1 = (ρ1)/2•V2•A2
Now, the pipe is uniform and thus the cross section doesn't change. Thus;
A1 = A2
We now have;
ρ1•V1•A1 = (ρ1)/2•V2•A1
A1 and ρ1 will cancel out to give;
V1 = (V2)/2
Thus, V2 = 2V1
