2 years ago

Risks in driving never begins with yourself, but with other drivers who take risks.

Engineering

1 answer:

Ymorist [56]2 years ago

8 0

False! Just saying. You could be under the influence, or just have no clue as to what you're doing.

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A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of

-Dominant- [34]

Answer:

a. $\tau=51.55 MPa$

b. $\tau=42.95MPa$

c. $\theta=7.67\times 10^{-3}$ rad.

Explanation:

Given: $D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa$

We know that

$\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}$

J for hollow shaft $J=\dfrac{\pi (D_o^4-D_i^4)}{64}$

(a)

Maximum shear stress $\tau =\dfrac{16T}{\pi Do^3(1-K^4)}$

$K=\dfrac{D_i}{D_o}$ ⇒K=0.83

$\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}$

$\tau=51.55 MPa$

(b)

We know that $\tau \alpha r$

So $\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}$

$\dfrac{51.55}{\tau}=\dfrac{210}{175}$

$\tau=42.95MPa$

(c)

$\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}$

$\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}$

$\theta=7.67\times 10^{-3}$ rad.

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4 years ago

A 1 phase load operates at 600 V and consumes 75 kW with a 0.85 lagging power factor. Compute the complex power consumed by the

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Answer:

Complex power=84 W

Explanation:

using equation

s=vs^2/2z

s=600^2/2z

s=84+29.4j

using s=P+jq

complex power=P=84 W

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4 years ago

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Answer:

C. exactly vertical

Explanation:

"Plumb" means "exactly vertical."

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The 5 ft wide gate ABC is hinged at C and contacts a smooth surface at A. If the specific weight of the water is 62.4 lb/ft3 , f

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Answer:

The solution is given in the attachments.

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to determine the rate of heat loss from that wall by convection = 12780 watts

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