The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water i
1 answer:
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Answer:
The shear stress will be 80 MPa
Explanation:
Here we have;
τ = (T·r)/J
For rectangular tube, we have;
Average shear stress given as follows;
Where;
= 100 mm × 200 mm = 20000 mm² = 0.02 m²
t = Thickness of the shaft in question = 2 mm = 0.002 m
T = Applied torque
Therefore, 50 MPa = T/(2×0.002×0.02)
T = 50 MPa × 0.00008 m³ = 4000 N·m
Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m
Therefore, = 0.05 m × 0.25 m = 0.0125 m².
Therefore, from the following average shear stress formula, we have;
Plugging in then values, gives;
The shear stress will be 80,000,000 Pa or 80 MPa.
Answer:
The resistance is 24.9 Ω
Explanation:
The resistivity is equal to:
The area is:
A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²
If NA is greater, then, the term 1/NA can be neglected, thus the equation:
Where
V = 0.44 V
E = 11.68*8.85x10¹⁴ f/cm
The length is:
L = 10 - 0.335 = 9.665 um
The resistance is:
Answer:
100 cm
Explanation:
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