a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:
where R is the radius of the wheel.
The period of revolution is:
Therefore, the tangential speed of the book is:
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
Answer:
a = the lowest critical speed of the shaft 882.81 rad/s
b = new diameter 0.05m or 50mm
c = critical speed 1765.62rad/s
Explanation:
see the attached file
The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.
Explanation:
The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.
As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.
Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.
Then ΔT = (70-(-10))=70+10=80°C.
As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.
Q = 1600C J
Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.
Answer:
The answer is "effective stress at point B is 7382 ksi
"
Explanation:
Calculating the value of Compressive Axial Stress:
![\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%20y%20%20%3D%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Cfrac%7B4%20F%7D%7B%28%20p%20d%20%5E2%20%29%7D%20%3D%20%5Cfrac%7B%284%20x%20%28%20-%2040000%20%5C%20lbf%29%29%7D%7B%5B%20p%20%5Ctimes%20%281%20%5C%20in%29%5E2%20%5D%7D%20%3D%20-%2050.9%20%5C%20ksi%20%5C%5C)
Calculating Shear Transverse:
![\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%27%20%3D%5B%20s%20y%5E2%20%2B3%28%20t%20%5Ctimes%20y%5E2%20%2B%20t%20yz%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C)
![= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi](https://tex.z-dn.net/?f=%3D%20%5B%20%28-50.9%29%5E2%20%2B3%28%2863.7%29%5E2%20%2B%280.17%29%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B%203%284057.69%29%2B0.0289%5D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B12%2C173.07%2B0.0289%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D14763.9089%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%20%3D%207381.95445%20%5C%20ksi%5C%5C%5C%5C%20%3D%207382%20%5C%20ksi)
