(a) Sabbir usually (sit)______ in the front bench.

How high of a column of sae 30 oil would be required to give the same pressure as 700 mm hg?

Rasek [7]

Hoiu-10,4000 mm.

<h3>Is positive pressure good for PC?</h3>

A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.

To learn more about it, refer

to https://brainly.in/question/413163

#SPJ4

7 0

1 year ago

Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes

Amanda [17]

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2: Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

5 0

3 years ago

Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur

Paha777 [63]

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.

(a). here we are asked to determine the Temperature and Pressure.

Given that the properties of Air;

ha = 230.02 KJ/Kg

Ta = 230 K

Pra = 0.5477

From the energy balance equation for a diffuser;

ha + Va²/2 = h₁ + V₁²/2

h₁ = ha + Va²/2 (where V₁²/2 = 0)

h₁ = 230.02 + 220²/2 ˣ 1/10³

h₁ = 254.22 KJ/Kg

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg

from this we have;

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]

Pr₁ = 0.77759

therefore T₁ = 254.15K

P₁ = (Pr₁/Pra)Pa

= 0.77759/0.5477 ˣ 26

P₁ = 36.91 kPa

now we calculate Pr₂

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349

⇒ now we obtain properties of air at

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg

calculating the enthalpy of air at state 2

ηc = h₁ - h₂ / h₁ - h₂

0.85 = 254.22 - 505.387 / 254.22 - h₂

h₂ = 549.71 KJ/Kg

to obtain the properties of air at h₂ = 549.71 KJ/Kg

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄ + 0 = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0

3 years ago

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,

Scorpion4ik [409]

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

4 0

3 years ago

A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate

olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

4 0

3 years ago