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Klio2033 [76]
2 years ago
11

Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.

Physics
1 answer:
Kamila [148]2 years ago
8 0

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

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Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi
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Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

  • Hf = head loss
  • f = friction factor
  • L = Length of the pipe = 360 m
  • V = Flow velocity, m/s
  • D = Pipe diameter = 0.12 m
  • g = Gravitational acceleration, m/s^2
  • Re = Reynolds's Number
  • rho = Density =998 kg/m^3
  • μ = Viscosity = 0.001 kg/m-s
  • Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

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2 years ago
What is the time required for an object starting at rest to fall freely 500 meters near earth surface
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Using kinematics, approximately 10.1 seconds
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5. How much does a suitcase weigh if it has a mass of 22.5 kg?
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27 pounds might be the answer cause 5.4 is 9 pound wick is an average size of an suite case
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A piston above a liquid in a closed container has an area of 1m2. The piston carries a load of 350 kg. What will be the external
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I say the answere would be A
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A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa
Alekssandra [29.7K]

Answer:

(a) 16.777mi

(b)Yes, he was speeding

Explanation:

(a)

Let's do the proper operations in order to convert km to mi:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

We can conclude that the trip length in miles was:

d=16.77706389mi

(b)

Let's calculate the speed of the man during the trip:

v=\frac{d}{t}

But first, let's do the proper operations in order to convert min to h:

16min*\frac{1h}{60min} =2.666666667h

Now, the speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

As we can see:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

So, we can conclude that the driver was speeding

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2 years ago
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