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Ray Of Light [21]
3 years ago
15

1. Which of the following system parts controls cool- ant flow through a radiator? (A) Fan (B) Radiator (C) Thermostat (D) Tempe

rature sensor, (8) prevent (C) prevent (O) control Supercharge cussed. Tech more oxygen​
Engineering
1 answer:
barxatty [35]3 years ago
7 0
Thermostat

Usually 160 or so degF
Water pump creates pressure differential/flow
Thermostat is closed only allowing minimum flow through a bypass hose. Water temp reaching set-point opens the thermostat for flow through the engine block and to radiator heat rejection.

Temperature sensor and electric 12vdc fan motor cycle fan on newer cars to control temperature but not flow.
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g Calculate a better value for the convection coefficient using resources from heat transfer. Assume forced convection of the ai
ollegr [7]

Answer:results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.

Explanation:

4 0
3 years ago
Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s
suter [353]

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

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3 0
2 years ago
the hoop is cast on the rough surface such that it has an angular velocity w=4rad/s and an angular acceleration a=5rad/s^2. also
IgorLugansk [536]

Given Information:

Angular velocity = ω = 4 rad/s

Angular acceleration = α = 5 rad/s²

Center deceleration = a₀ = 2 m/s

Required Information:

Acceleration of point A at this instant = ?

Answer:

Acceleration of point A at this instant = 5.94 m/s²

Explanation:

Refer to the attached diagram of the question,

The acceleration of point A is given by

a = a₀ + rα - rω²

Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.

a = -2i + 0.3j*5k - 0.3j*4²

a = -2i + 1.5(j*k) - 0.3j*16

a = -2i + 1.5(-i) - 4.8j

a = -2i - 1.5i - 4.8j

a = -3.5i - 4.8j

The magnitude of acceleration vector is

a = √(-3.5)² + (-4.8)²

a = √35.29

a = 5.94 m/s²

Therefore, the acceleration of point A is 5.94 m/s²

The angle is given by

θ = tan⁻¹(y/x)

θ = tan⁻¹(-4.8/-3.5)

θ = 53.9°

7 0
4 years ago
Technician A says you should place the air ratchet setting to
Bas_tet [7]
I think it’s b sry if I’m wrong tho
6 0
3 years ago
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
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