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Alexeev081 [22]

3 years ago

8

A rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind

tunnel. The freestream velocity is 40 m/s. The density of the air is 1.2250 kg/m3, and the absolute viscosity is 1.7894 x 10-5 kg/m s. Plot (not sketch) the velocity profiles at x = 0.0 m, x = 0.05 m, x = 0.10 m, and x = 0.20 m. Calculate the chordwise distribution of the skin friction coefficient and the displacement thickness. What is the drag coefficient for the plate? Be sure to account for both sides of the plate.

1 answer:

34kurt3 years ago

4 0

Answer:

Answer for the question is given in the attachment.

Explanation:

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Please select the word from the list that best fits the definition a puffy white cloud with a flat bottom

ankoles [38]

Cumulus? They are fluffy clouds that have flat bottoms. Is this what you need?

3 0

4 years ago

The shoe is in contact with the initially nearly stationary 0.500 kg football for 20.0 ms. What average force is exerted on the

Olin [163]

If v=22m/s after0.02s then acceleration was
22/0.02=1100m/s^2
Now, F=ma=(0.5)(1100)=550N

4 0

3 years ago

Read 2 more answers

Please help 50 points

PilotLPTM [1.2K]

Answer:

3a 0.7m/s

3b partially inelastic

4a 8.33 m/s

4b completely inelastic

5. puck: 10.59 m/s octopus: 10.59m/s

6. car: 117.44 m/s truck: 17.44m/s

7a - 1 m/s , the red cart travels to the left

7b) elastic

Explanation:

For all these questions, you have find momentum (P=mv) always remember initial P is always equal to final P

3.

Initial P:

mass of first ball x velocity of first ball + mass of second ball x velocity of second ball

0.5*3.5 + 0.5*0 = 1.75 kg.m/s

final P: also 1.75Kg.m/s

let x be the velocity for first ball

0.5*2.8+0.5*x=1.75

0.5x=1.75-1.4

x=0.35/0.5

x=0.7m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is partially inelastic, partially because the ball did not stick together

4.

Initial P:

2575*11 + 825*0 = 28,325 kg.m/s

final P: also 28,325 Kg.m/s

let x be the velocity for both vehicles

(2575+825)x=28325

x=28325/3400

x=8.33 m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is completely inelastic, completely because the vechicles slides off together

5

Initial P:

0.115*35 + 0.265*0 = 4.025 kg.m/s

final P: also 4.025Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the puck and octopus,

(0.115+0.265)x=4.025

x=4.025/0.38

x=10.59

6. Initial P:

565*25+785*12 = 23,545 kg.m/s

final P: also 23,545 Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the car and truck,

(565+785)x=23545

x=23545/1350

x=17.44

7.

Initial P:

0.25*2 + 0.75*0 = 0.5 kg.m/s

final P: also 0.5 Kg.m/s

let x be the velocity for red cart

0.75*1+0.25*x=0.5

0.25x=0.5-0.75

x=-0.25/0.25

x=-1m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi equals to KEf

so the collision is elastic

7 0

3 years ago

Sonja [21]

Answer:

The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. ... These sounds will interfere to produce detectable beats.

Explanation:

5 0

3 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

4 0

3 years ago