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lakkis [162]
2 years ago
9

A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside

diameter of a hole as its temperature increases
Engineering
1 answer:
djverab [1.8K]2 years ago
4 0

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

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A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

6 0
3 years ago
You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.
allochka39001 [22]
Answer:








Explanation:









I hope this helps!
3 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
Question 74
torisob [31]

Answer:

C). rearview mirror (at least of four inches by four inches).

Explanation:

<u>A 'rearview mirror' would be needed in case one is required to pull an individual on water skis after a PWC(personal watercraft) that is rated for carrying two persons only</u>. This wide-angle mirror for rearview would allow the operator to monitor the person who is being towed constantly along with riding the ski at the same time. It is considered illegal to tow a person on a vessel without having a rearview mirror and at the same time, the limited capacity must also be followed strictly. Hence, <u>option C</u> is the correct answer.

7 0
3 years ago
You work in a furniture store. You receive a
spin [16.1K]

Answer:

18 pieces of furniture

Explanation:

Since you receive $120.93 per furniture piece and a the month's commission is $2,176.74 you divide the commission by the furniture price.

  • 2176.74/120.93
3 0
3 years ago
Read 2 more answers
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