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3 years ago

Three-dimensional measuring references all of these EXCEPT:

Engineering

1 answer:

cricket20 [7]3 years ago

4 0

Except the Table of Contents

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Question 1

vlabodo [156]

Answer:

A. from zero at the center to maximum at the circumference.

Explanation:

The shear stress describes the process where a body deforms as a result of a part sliding over the other.

The value of shear stress which is induced in the shaft due to the applied couple varies from zero at the center to maximum at the circumference.

The shear stress in a shaft is given by this formula;

τ = Tr/I

where;

τ = shear stress (MPa).

T = applied torque (Nmm).

r = distance from center to stressed surface in the given position (mm).

I = polar moment of Inertia of cross-sectional area (mm).

3 0

3 years ago

Cell phones require powerful batteries in orde to work effectively. Which activity is best described as an engineering endeavor

natta225 [31]

Where are the options?

8 0

3 years ago

How to find Catenary length in sag

Gnom [1K]

Answer:

buter uwu heheheheh joke

6 0

2 years ago

In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets

Whitepunk [10]

Answer:

The answer is VN =37.416 m/s

Explanation:

Recall that:

Pressure (atmospheric) = 100 kPa

So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles

Now,

Pabs =Patm + Pgauge = 800 KN/m²

Thus

PT/9.81 + VT²/2g =PN/9.81 + VN²/2g

Here

Acceleration due to gravity = 9.81 m/s

800/9.81 + 0

= 100/9.81 + VN²/19.62

Here,

9.81 * 2= 19.62

Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

7 0

3 years ago

Read 2 more answers

An automobile having a mass of 1100 kg initially moves along a level highway at 120 km/h relative to the highway. It then climbs

soldier1979 [14.2K]

Answer:

$-6111.11\ \text{kJ}$

$863.28\ \text{kJ}$

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = $\dfrac{120}{3.6}\ \text{m/s}$

h = Height of hill = 80 m

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Change in kinetic energy

$KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}$

Change in kinetic energy is $-6111.11\ \text{kJ}$

Change in potential energy is given by

$PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}$

The change in potential energy is $863.28\ \text{kJ}$ .

8 0

3 years ago