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2 years ago

What is the primary damage control telephone circuit for

Engineering

1 answer:

user100 [1]2 years ago

6 0

Answer:

hsyghcjqg9ug9duyssatayfjzurldh

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Only put coolant into your radiator when the engine is _____.

Arada [10]

Only put ciilant into ur radiator when the engine is cool (D)

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3 years ago

⦁ An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and

Anon25 [30]

Answer:

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8 0

4 years ago

Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

Nina [5.8K]

dutile is the correct answer

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3 years ago

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

olga55 [171]

Answer:

$P_2_{abs}=160\ psia$ (absolute).

Explanation:

Given that

Pressure ratio r

r=8

$r=\dfrac{P_2_{abs}}{P_1_{abs}}$

$8=\dfrac{P_2_{abs}}{P_1_{abs}}$ -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure + Gauge pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5 psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

$8=\dfrac{P_2_{abs}}{20}$

$P_2_{abs}=8\times 20\ psia$

$P_2_{abs}=160\ psia$

Therefore the exit gas pressure will be 160 psia (absolute).

7 0

3 years ago

Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

6 0

3 years ago