Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s
Hello!
Since the two weights are <em>off</em> the table, the block will move towards letter F.
I hope this helps :))
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
2 J
Explanation:
A charged capacitor of capacitance 
with energy of 7.54 J, is connected in parallel with another capacitor 
, so the charge is equally distributed between them.
(a) The energy stored in the capacitor before it being connected to the other capacitor is:
The energy stored in the electric field is the sum of the energies of the two capacitors:
since the charge equally distributed, 
= 
= 
. and since they are connected in parallel the potential difference on both of them is the same
:
hence,
