Answer:
<em>No, the velocity profile does not change in the flow direction.</em>
Explanation:
In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>
Answer:
Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...
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Suzuki Hayabusa – top speed: 248 mph. 1340cc
Answer:
A) About 
newtons
B) 76.518 newtons
C) 111.834 newtons
Explanation:
A) 
, where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:
B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.
C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.
Hope this helps!
Answer:
True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.
Option: A
<u>Explanation:
</u>
To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. <em>Regeneration</em> technique <em>is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine. </em>
<em>Thermal efficiency</em> of a turbine is increased as <em>the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.
</em>
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius, 
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax 
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
