We have that for the Question, it can be said that the amount by which the length of the stack decreases is
From the question we are told
A copper (<em>Young's modulus </em>1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each <em>cylinder </em>has a radius of 0.24 cm.
A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack <em>decreases</em>.
Generally the equation for <em>copper </em>cylinder is mathematically given as
Generally the equation for brass<em> </em>cylinder is mathematically given as
Therefore Total change in length
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Answer:
0·95
Explanation:
Given the combined mass of the rider and the bike = 100 kg
Percent slope = 12%
∴ Slope = 0·12
Terminal speed = 15 m/s
Frontal area = 0·9 m²
Let the slope angle be β
tanβ = 0·12
As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions
The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached
From the diagram m × g × sinβ = drag force
Drag force = 0·5 × d × 
× v² × A
where d is the density of the fluid through which it flows
is the drag coefficient
v is the speed of the object relative to the fluid
A is the cross sectional area
As tanβ = 0·12
∴ sinβ = 0·119
Let the fluid in this case be air and density of air d = 1·21 kg/m³
m × g × sinβ = 0·5 × d × × v² × A
100 × 9·8 ×0·119 = 0·5 × 1·21 × × 15² × 0·9
∴ ≈ 0·95
∴ Drag coefficient is approximately 0·95
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Answer:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
