Answer:
<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>
Explanation:
The question is incomplete. So the picture, which is missing in question, is attached for your review.
As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).
<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>
Answer: the increase in the external resistor will affect and decrease the current in the circuit.
Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is
E = IR + Ir = I(R + r)........(1)
Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)
I = E/(R + r)
As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.
Answer:
The answer is below
Explanation:
a) The weight of the combined system is the sum of the weight of the water and the weight of the tank
b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.
c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.
d) Given that:
Answer:
c. V2 equals V1
Explanation:
We can answer this question by using the continuity equation, which states that:
(1)
where
A1 is the cross-sectional area in the first section of the pipe
A2 is the cross-sectional area in the second section of the pipe
v1 is the velocity of the fluid in the first section of the pipe
v2 is the velocity of the fluid in the second section of the pipe
In this problem, we are told that the pipe has a uniform cross sectional area, so:
A1 = A2
As a consequence, according to eq.(1), this means that
v1 = v2
so, the velocity of the fluid in the pipe does not change.
Answer:
The average velocity of the needle exit flow is 163.27 m/s
Explanation:
Given;
inside diameter of the syringe, d₁ = 20 mm
inside diameter of the needle, d₂ = 0.7 mm
steady flow rate of the plunger, v = 20 mm/s
Apply continuity equation;
Considering 0.1 of the volume flowrate out the needle opening, we will have the following equations.
Therefore, the average velocity of the needle exit flow is 163.27 m/s
