Answer:
386.6 A
Explanation:
We are given;
Solenoid length; L = 10 cm = 0.1 m
Number of turns; N = 247
Magnetic field; B = 1.2 T
Now, at the centre of a long solenoid of N turns/metre carrying a current(I), the formula for the magnetic field is given as;
B = μ_oNI/L
Making I the subject we have;
I = BL/μ_oN
Where μ_o is the vacuum of permeability and has a constant value of 4π × 10^(−7) H/m
Thus;
I = (1.2 × 0.1)/(4π × 10^(−7) × 247)
I = 386.6 A
An object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of θ with the horizontal.
1. Which direction does the force of gravity act on the object?
2. Which direction does the normal force act on the object?
Which force is responsible for the object moving down the incline?
Answer:
The answer is below
Explanation:
1. When an object is moving with a constant velocity, the direction the force of gravity act on the object is DIRECTLY DOWN.
2. When an object is moving with a constant velocity, the direction the normal force act on the object "perpendicular to the surface of the plane."
3. When an object is moving with a constant velocity, the force that is responsible for the object moving down the incline is "the component of the gravitational force parallel to the surface of the inclined plane."
Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
The important point here is that volumetric flow rate in the pump and the pipe is the same.
Q = AV, where Q = Volumetric flow rate, A = Cross sectional area, V = velocity
Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V
Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s
Therefore, the flow of water in the pipe is 5000 cm/s.
