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3 years ago

The B-pillar may also be called the:

Engineering

1 answer:

slega [8]3 years ago

6 0

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up

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Convert 250 lb·ft to N.m. Express your answer using three significant figures.

vfiekz [6]

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So $250lb-ft=250\times 4.45N\times 0.348M=338.95N-m$

So 250 lb-ft = 338.95 N-m

6 0

3 years ago

I need help with this question

Ad libitum [116K]

Answer:

LOL where is the question, that u need help with?

Explanation:

5 0

3 years ago

Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i

Brilliant_brown [7]

Answer:

$\dot Q_{in} = 372.239\,MW$

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

$w_{in} + h_{in}- h_{out} = 0$

$h_{out} = w_{in}+h_{in}$

$h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}$

$h_{out} = 203.81\,\frac{kJ}{kg}$

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

$h_{out} = 2685.4\,\frac{kJ}{kg}$

The rate of heat transfer in the boiler is:

$\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)$

$\dot Q_{in} = 372.239\,MW$

3 0

3 years ago

Read 2 more answers

The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id

baherus [9]

Answer:

attached below

Explanation:

3 0

3 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

jok3333 [9.3K]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$ .

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

We know that

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where Q is volume flow rate

L is length of tube

$d_i$ is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$ .

6 0

3 years ago