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Stolb23 [73]
3 years ago
11

I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!

Engineering
2 answers:
allsm [11]3 years ago
6 0

Answer:

the answer how you analyzs the problwm

romanna [79]3 years ago
4 0

Answer:

To access a missions, simply log on to your CoderZ account and navigate to one of the courses you are assigned, choose a pack, and then choose one of the missions!

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A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu
Mumz [18]

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

Density = \frac{mass}{volume}

Therefore, mass is:

mass = denisty *volume

Density of air = 2 kg/m^3; Volume of the tank =  3.5 m^3

=> Mass = 3.5 * 2 = 7 kg

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = 6.5 kg/m^3

The new mass will be:

Mass = 6.5 * 3.5 = 22.75 kg

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

6 0
3 years ago
For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin
Ad libitum [116K]

Answer:The awnser is 5

Explanation:Just divide all of it

4 0
3 years ago
A cyclic tensile load ranging from 0 kN to 55 kN force is applied along the length of a 100 mm long bar with a 15 mm x 15 mm squ
Yuliya22 [10]

Answer:

square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.

Required:

a. What is the nominal maximum tensile stress on the bar?

b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?

7 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
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