Can you help me with this task/homework.

The first step to merging is entering the ramp and _____.

Yuri [45]

Answer:

D.telling your passengers where you are going

7 0

3 years ago

What is the answer???

coldgirl [10]

Answer:

C IS THE ANSWER

Explanation:

HOPE IT HELPS :))))))))) <3

5 0

3 years ago

Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?

zhenek [66]

Answer:

it is indeed C

Explanation:

4 0

3 years ago

Read 2 more answers

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)

mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

$q=1.7845$

Explanation:

From the question we are told that:

Air Temperature $T_1=40c$

Length $l=2m$

Velocity $v=7m/s$

Width $w=0.5$

Constant temperature $T_t= 100C$

Generally the equation for Total heat Transfer is mathematically given by

$q=hA(T_s-T_\infty)$

Where

h=Convective heat transfer coefficient

$h=29.9075w/m^2k$

Therefore

$q=h(L*B)(T_s-T_\infty)$

$q=29.9075*(2*0.5)(100+273-(40+273))$

$q=1794.45w$

$q=1.7845$

5 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago