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3 years ago

8

A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False

)

1 answer:

tiny-mole [99]3 years ago

4 0

The answer is : True

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A(n) _____ is an apparatus that changes alternating current (AC) to direct current (DC)

amid [387]

Answer:

rectifier

Explanation:

7 0

2 years ago

Can someone help me plz!!

pogonyaev

It has to do with mechanical engineering

6 0

3 years ago

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

$v^2-u^2=2as$

where,

s = distance covered by the object = 6.93 m

u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

7 0

3 years ago

A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of

zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

So Total net Force on each bolt Fb = P_Eb + Pi

Fb = 4.9 kip + 40 kip

Fb = 44.9 kip

Therefore the net force in every bolt is 44.9 kip

4 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