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2 years ago

An urgent. Please answer this. :)1 & 2 are Ω3 & 4 are V

Engineering

1 answer:

balandron [24]2 years ago

6 0

Answer:

See below ↓

Explanation:

1) 40 Ω

2) 24 Ω

3) 85 V

4) 135 V

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A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.

ddd [48]

Answer:

$P = 150.335\,kPa$ (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 150.335\,kPa$

4 0

3 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

2 years ago

Each of the following activities are commonly performed during the implementation of the Database Life Cycle (DBLC). Fill in the

kicyunya [14]

Yessiree I agree with yu cause yu are right

4 0

2 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

2 years ago

Which of the following is typically wom when working in an atmosphere containing dust?

alukav5142 [94]

Answer:

Either D or C

Both of these masks are used for dust, but since half masks are generally cheaper and easier to use, I'd go with C.

If this is correct, I'd appreciate a brainliest.

3 0

2 years ago

An urgent. Please answer this. :)1 & 2 are Ω3 & 4 are V​

An urgent. Please answer this. :)1 & 2 are Ω3 & 4 are V