Forces always act in equal and opposite pairs

ELEN [110]

Answer:Ballydoyle is a racehorse training facility located in County Tipperary in Ireland. It is a sister thoroughbred facility to Coolmore Stud, and both are owned by John Magnier, son in law to the racehorse trainer Vincent O'Brien.

Explanation:

5 0

4 years ago

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A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a stat

dalvyx [7]

Answer:

123.9 Btu

Explanation:

The energy balance on the air is:

∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W

ignore ∆KE and ∆PE,

W = ∆U − Q = m(u2 − u1) − Q; (u2 − u1 = 51.94 Btu/lb)

ideal gas properties is attached

W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu

u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb

6 0

3 years ago

What is the most likely reason the rover won't travel in a straight line?

vovangra [49]

The rovers were designed to trek up to 100 meters (about 110 yards or 328 feet) across the martian surface each martian day, though they have gone much farther. While a complete martian day (called a sol) is about 24 hours and 40 minutes long (or 24 hours 37.5 minutes if you prefer), the Sun can only provide enough power for driving during a four-hour window around high noon. That means the rovers have to be able to move quickly and effectively.
Moving safely from rock to rock or location to location is a major challenge because of the communication time delay between Earth and Mars, which is about 20 minutes on average. Unlike a remote controlled car, the drivers of rovers on Mars cannot instantly see what is happening to a rover at any given moment and they cannot send quick commands to prevent the rover from running into a rock or falling off of a cliff.
During surface operations on Mars, each rover receives a new set of instructions at the beginning of each sol. Sent from the scientists and engineers on Earth, the command sequence tells the rover what targets to go to and what science experiments to perform on Mars. The rover is expected to move over a given distance, precisely position itself with respect to a target, and deploy its instruments to take close-up pictures and analyze the minerals or elements of rocks and soil.

6 0

3 years ago

Cutting and forming operations performed on relatively thin sheets of metal. a)- True b)-False

belka [17]

Answer: True

Explanation: There are different actions that are performed on the metal sheet . Cutting operation is the operation of splitting,slitting or parting of the metal sheet on the by using different kind of blades . Formation operation are basically for the exerting pressure on the sheet to give it desired shape such as bending, squeezing etc. Therefore the given statement is true.

6 0

3 years ago