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Verdich [7]
3 years ago
6

At which minimum height off the floor should a new oven be installed?

Engineering
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

At 30mm

Explanation:

hope it helps

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An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40
Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0
3 years ago
Which statement explains what causes the balloon to accelerate
victus00 [196]
The air leaving through the balloon's mouth pulls the balloon in the same direction as the exiting air, so the balloon experiences a net force. All air surrounding the balloon pushes the balloon forward.
5 0
3 years ago
What is the main reason for using a complex data table?
wolverine [178]

Answer:

D. To interpret the possible meaning of the data

7 0
3 years ago
Read 2 more answers
Consider a piston-cylinder device with a piston surface area of 0.1 m^2 initially filled with 0.05 m^3 of saturated water vapor
miv72 [106K]

The friction force f = 10000 N

The heat transfer Q = 1.7936 KJ

<u>Explanation:</u>

Given data:

Surface area of Piston = 1 m^{2}

Volume of saturated water vapor = 100 K Pa

Steam volume = 0.05 m^{3}

Using the table of steam at 100 K pa

Steam density = 0.590 Kg/m^{3}

Specific heat C_{p} = 2.0267 KJ/Kg K

Mass of vapor = S × V

m = 0.590 × 0.05

m = 0.0295 Kg

Solution:

a) The friction force is calculated

Friction force = In the given situation, the force need to stuck the piston.

= pressure inside the cylinder × piston area

= 100 × 10^{3}  × 0.1

f = 10000 N

b)  To calculate heat transfer.

Heat transfer = Heat needs drop temperatures 30°C.

Q=m c_{p} \ DT

Q = 0.0295 × 2.0267 × 10^{3} × 30

Q = 1.7936 KJ

3 0
4 years ago
Shear plane angle and shear strain: In an orthogonal cutting operation, the tool has a rake angle = 16°. The chip thickness befo
Oduvanchick [21]

Answer:

shear plane angle Ф = 26.28°

shear strain 2.20

Explanation:

given data

angle = 16°

chip thickness t1 = 0.32 mm

cut yields chip thickness t2 = 0.72 mm

solution

we get here first chip thickness ratio that is

chip thickness ratio = \frac{t1}{t2}    ................. 1

put here value

chip thickness ratio  = \frac{0.32}{0.72}  

chip thickness ratio r = 0.45

so here shear angle will be Ф

tan Ф = \frac{r*cos\alpha }{1-rsin\alpha}   ............2

tan Ф = \frac{0.45*cos16 }{1-rsin16}  

tan Ф = 0.4938

Ф = 26.28°

and

now we get shear strain that is

shear strain r = cot Ф + tan (Ф - α )   ................3

shear strain r  = cot(26.28) + tan (26.28 - 16 )

shear strain r = 2.20

6 0
4 years ago
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