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kobusy [5.1K]
3 years ago
7

Differnence between boat and ship​

Engineering
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

there you go

Explanation:

With regard to motorized craft, a ship is a large vessel intended for oceangoing or at least deep-water transport, and a boat is anything else." Basically, a ship can carry a boat, but a boat cannot carry a ship

Blababa [14]3 years ago
5 0

Answer:

a ship is a large vessel intended for oceangoing or at least deep-water transport, and a boat is anything else." Basically, a ship can carry a boat, but a boat cannot carry a ship

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a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced
ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0
2 years ago
Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid wat
Evgen [1.6K]

Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

For steam,

m_{s}=5.55 \ kg/s

\Delta h=2491 \ kg/kj

For water,

\Delta T=10^{\circ}C

(Cp)_{w}=4.184 \ kJ/kg^{\circ}C

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒  m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T

On putting the estimated values, we get

⇒  5.55\times 2491=M_{w}\times 4.184\times 10\\

⇒  13825.05=M_{w}\times 41.84

⇒  M_{w}=330 \ kg/s

7 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w
shusha [124]

Answer:

Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.

5 0
3 years ago
PLEASE HURRY!!!
Naily [24]

Answer:

A

Explanation:

He should get a job in engineering to see what it's like to work in the field.

3 0
2 years ago
Read 2 more answers
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