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Anna007 [38]
3 years ago
13

The tolerance for a geometric can be easily identified by reading the _________.

Engineering
2 answers:
notsponge [240]3 years ago
6 0
The answer is dimension
baherus [9]3 years ago
4 0

Answer:

C

Explanation:

You might be interested in
A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.
ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

  D=200 mm

 P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

       hoop stress \sigma _{h}=\frac{Pd}{2t}

      Longitudinal stress \sigma _{l}=\frac{Pd}{4t}

So maximum shear tress in plane\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}

              \tau _{max}=\dfrac{Pd}{8t}

Now by putting the value

       27.5=\dfrac{4\times 200}{8t}

 So   t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
3 years ago
A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa
Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

          Energy produce in one year = 204984 x 10¹² J/year

          Energy produce in one year =20.49 x 10¹⁶ J/year

7 0
4 years ago
For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste
andrew11 [14]

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

3 0
3 years ago
No question but thx<br> reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
jeka94

Answer:

why you doin this

Explanation:

is this so we get free points?

5 0
3 years ago
Read 2 more answers
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