Ask question

Ask question

All categories

Misha Larkins [42]

3 years ago

6

1. Describe simply what will happen to an airplane in flight in the following conditions:

1 answer:

notka56 [123]3 years ago

4 0

Answer:

For the two you haven't answered: (Drag greater than thrust, lift greater than weight) It will accelerate backwards (decelerate) and upwards

(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.

You might be interested in

Tech A says that it is best to use a knife or other type of sharp tool to cut away the insulation when

Marina86 [1]

Tech A because it is best to use a knife

3 0

3 years ago

You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

allochka39001 [22]

Answer:

For most uses you'll want your water heated to 120 F(49 C) In this example you'd need a demand water heater that produces a temperature rise and it will take about 2 hours

7 0

3 years ago

View the picture below and then correctly answer the questions using the following words: Temperate Zone, Tropical Zone, Polar Z

Nadusha1986 [10]

Can you provide the picture? Thanks !

8 0

3 years ago

Read 2 more answers

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t

HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid F

$F=\rho AV^2$

So force exerted in x-direction

$F_x=\rho AV^2$

$F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So force exerted in y-direction

$F_y=\rho AV^2$

$F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So the resultant force R

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{0.763^2+0.763^2}$

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0

3 years ago