Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A = 
= 
Momentum of block B = 
= 
After collision:
Momentum of block A = 
= 
Applying law of conservation of momentum to find momentum of block B after collision 
.
Plugging in the given values and simplifying.
Adding 200 to both sides.
∴ 
Momentum of block B after collision =
Current = (voltage) / (resistance)
= (35 volts) / (1,400 ohms)
= 25 milliamperes
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
100/5= 20, 20 seconds is the time needed.
Answer:
option C
Explanation:
the ball is moving circular around the pole
Angular momentum of the system is constant
J = I ω
now,
the rope radius is decreasing as it revolving around the pole
angular speed is inversely proportional to radius.
so, the angular speed will increase.
The correct answer is option C
