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irakobra [83]
3 years ago
7

Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has

a small hole through which water leaks out at a rate of 0.03 kg / s. If the tank initially contained 40 kg of water, how much will it have after 2 minutes?
Engineering
1 answer:
stiks02 [169]3 years ago
8 0

Answer:

total amount of water after 2 min will be 84.4 kg/s

Explanation:

Given data:

one tank inflow = 0.1 kg/s

2nd tank inflow = 0.3 kg/s

3rd tank outflow = 0.03 kg/s

Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s

From third point, outflow is 0.03 kg/s

Therefore, resultant in- flow = 0.4 - 0.03

Resultant inflow is  = 0.37 kg/s

Tank has initially 40 kg water

In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg

So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg

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390.242 MPa

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2 years ago
7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
Crank

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

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From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

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Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
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Answer:

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answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

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x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

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