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bixtya [17]
3 years ago
11

You want a potof water to boil at 105 celcius. How heavy a

Engineering
1 answer:
ankoles [38]3 years ago
3 0

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

area = \frac{\pi }{4} 0.15^2

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = \frac{Fnet}{g}

mass  = \frac{350}{9.8}

mass = 35.7 kg

so 35.7 kg lid we put

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You are hired as the investigators to identify the root cause and describe what should have occurred based on the following info
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Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

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To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' =  803 kg/m³ × 4.916 m³ = 3947.548 kg

So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg

<u>Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.</u>

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3 years ago
If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

                                            =0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

note:

find the attached pic

7 0
3 years ago
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