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Eva8 [605]
4 years ago
5

The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

ear with depth what is the surface velocity of flow in the channel?
Engineering
1 answer:
IceJOKER [234]4 years ago
4 0

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

\dot{m}=\rho AV

Putting all the value to get the velocity of the flow

\frac{\dot{m}}{\rho A} = V

V = \frac{4000}{1000*4*2}

V = 0.5 m/s

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Determine ten different beam loading values that will be used in lab to end load a cantilever beam using weights. Load values sh
nasty-shy [4]

Answer:

1st value = 1.828 * 10 ^9 gm/m^2 -------     10th value = 7.312 * 10^9 gm/m^2

Explanation:

initial load ( Wp) = 200 g

W1 ( value by which load values increase ) = 100 g

Ten different beam loading values :

Wp + w1 = 300g ----- p1

Wp + 2W1 = 400g ---- p2

Wp + 3W1 = 500g ----- p3 ----------------- Wp + 10W1 = 1200g ---- p10

x = 10.25" = 0.26 m

b = 1.0" = 0.0254 m

t = 0.125" = 3.175 * 10^-3 m

using the following value to determine the load values at different beam loading values

attached below is the remaining part fo the solution

5 0
3 years ago
What are some homophones​
ryzh [129]

Answer:

accessary, accessory.

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air, heir.

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all, awl.

allowed, aloud.

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5 0
3 years ago
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I want to solve the question
DedPeter [7]

Answer:

yes.

Explanation:

5 0
3 years ago
Differentiate between "Threshold and Resolution" with suitable examples.
9966 [12]

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

6 0
3 years ago
A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l
arlik [135]

Answer:

The displacement from t = 0 to t = 10 s,  is -880 m

Distance is 912 m

Explanation:

v = (12 - 3t^2) m/s = ds/dt.  .  . . . . . . . .  A

integrate above equation we get

s = 12t - t^3 + C

from information given in the question  we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

s(t) = 12t - t^3 - 21

we know that acceleration is given as

a(t) = dv/dt = -6t  

[FROM EQUATION A]

Acceleration at  t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m

the distance the particle travels during this time period:

let v = 0,

3t^2 = 12

t = 2 s

Distance = [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m

7 0
4 years ago
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