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Hatshy [7]
3 years ago
15

Which feature would be used to measure OUTER diameter?

Engineering
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

Calipers

Explanation:

Calipers can also be used to measure holes of different shapes: square, rectangular, cylindrical or hexagonal. The lower set of jaws on the caliper is used to measure the outer diameter of a cylinder. It will also give you the total length or width of an object.

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What factors determine tire inflation pressure?
Virty [35]

Answer:

Tire inflation can be caused by temperature and speed :)

6 0
3 years ago
14. Tires are rotated to
ExtremeBDS [4]

Answer:

A

Explanation:

3 0
3 years ago
Read 2 more answers
A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t
iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, P_{1}= 1 bar

v_{2} = 4 bar

pV¹°¹ = constant

From superheated propane table, at  P_{1}= 1 bar andT_{1}  = 27⁰C

v_{1} = 0.557 m³/kg

v_{2} = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1  \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557  \\\\V2 = 0.158 m^3/kg

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\  \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\   \\W = -19.75KJ

(b) Heat transfer = ?

Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ

8 0
4 years ago
Why is it reasonable to say that no system is 100% efficient?​
Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

5 0
3 years ago
3 cm of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at
ycow [4]

Answer:

water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

Explanation:

Given the data in the question;

A = 200 hectare =  2 × 10⁶ m²

water evaporated Ve = 2 × 10⁶ m² × 3 × 10⁻² = 60000 m³ { in 24 hrs }

Water added by storm in 24hrs Vi = 3 × 24 × 3600 = 259200 m³

now let water released be Vr

ΔV = V_ini - V_final = 0

Vi - Ve - Vr = 0

Vr = Vi - Ve

Vr = 259200 m³ - 60000 m³

Vr = 199200 m³ = 19920000 m² - cm

Vr = 1992 ha-cm

Therefore, water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

7 0
3 years ago
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