1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
professor190 [17]
3 years ago
15

Hey answr this sajida Yusof

Engineering
2 answers:
Varvara68 [4.7K]3 years ago
7 0

Answer:

that not even a question

Allisa [31]3 years ago
5 0
What do u need? Rusbaisuwvwbs
You might be interested in
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
Read 2 more answers
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
3 years ago
failure to yield the right-of-way to another vehicle or pedestrian is the primary Collision factor in about 20% of fatal and inj
laila [671]

You could just create a turn on green arrow. Have a button for pedestrians. The only way for them to turn is if they have a green arrow and the green arrow will only appear when pedestrians are stopped or finished walking and all cars are clear or fully stopped

8 0
4 years ago
A clutch is chattering. Technician A says that uneven wear on the flywheel could be the cause. Technician B says that worn or br
svet-max [94.6K]

Answer:

Both Technician A and B are correct

Explanation:

Due to a contamination of the clutch disc friction surface, the clutch chatters

The contamination that causes clutch chattering includes

1) Worn or broken motor mounts

2) Bell housing bolts becoming loose

3) Clutch link damage

4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering

5) Engine or transmission oil contaminating the disc.

4 0
3 years ago
The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t
tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

\dfrac{dv}{dt}=a

\dfrac{dv}{dt}=(2t-10)

integrating both side

\int dv =\int (2t -10) dt

 v = t² - 10 t + C

at t = 0   v = 3

so, 3 = 0 - 0 + C

     C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

\dfrac{dx}{dt}=v

\dfrac{dx}{dt}=(t^2-10t + 3)

integrating both side

\int dx =\int (t^2-10t + 3)dt

x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C

now, at t= 0 s = -4

-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C

C = -4

So,

x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4

Position function is equal to x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4

8 0
3 years ago
Other questions:
  • Joseph wants to practice architecture. Which compulsory assessment administered by NCARB does he need to complete?
    10·1 answer
  • The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associati
    10·1 answer
  • A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the
    10·1 answer
  • Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address
    14·1 answer
  • You can safely place a jack on a floor pan to keep a vehicle steady.
    5·2 answers
  • The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and
    15·1 answer
  • In your role, you are responsible for making sure that delivery trucks depart from the building on time. When a truck arrives la
    10·1 answer
  • 28. What is the value of a resistor in a series circuit if you measure 0.5 amps flowing through it and 15 volts
    10·1 answer
  • Explain the problems and their possible solution for electricity problems ?​
    8·1 answer
  • Whose responsibility is it to provide direction on correct ladder usage?<br> select the best option.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!