Answer:
Total Work=2275000 ft-lb
Explanation:
According to Riemann sum approximate for work needed to lift the cable:
Sine we have to add 650 terms because distance 650 ft, we will us the integration.
![W=\int\limits^a_b {f(x)} \, dx \\W=\int\limits^{650}_0 {8x} \, dx \\W=[4x^2]_0^{650}\\W=4(650)^2-0^2\\W=4*422500 ft-lb\\W=1690000 ft-lb](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5Ea_b%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B650%7D_0%20%7B8x%7D%20%5C%2C%20dx%20%5C%5CW%3D%5B4x%5E2%5D_0%5E%7B650%7D%5C%5CW%3D4%28650%29%5E2-0%5E2%5C%5CW%3D4%2A422500%20ft-lb%5C%5CW%3D1690000%20ft-lb)
Work done on lifting:
Total Work= 
Total Work=1690000+585000
Total Work=2275000 ft-lb
<span>An object that is in free fall seems to be
weightless, even though the
usual force of gravity is acting on it.
If the falling object happens to be a person, then every part of him is
falling at the same rate, so none of his internal organs are pressing
against anything like they normally do, and he has no sensation of weight.
If a bathroom scale happens to be falling with him and he places it under
his feet, it won't read anything, since the scale and the person are both
falling at the same rate. </span>
The particle with sharp ends have the slowest rate of deposition
Answer: Option C
<u>Explanation:</u>
As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.
Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.
Answer:
1.35 A
Explanation:
Applying,
V = IR
I = V/R..................... Equation 1
I = Current, V = Voltage, R = Resistance.
But,
R = Lρ/A............... Equation 2
Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.
Sustitute equation 2 into equation 1
V = AV/Lρ............... Equation 3
From the question,
Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m
Substitute these values into equation 3
I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )
I = (2.03×10⁻⁷)/(15×10⁻⁸)
I = 1.35 A
