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dimulka [17.4K]
3 years ago
13

Water flows at 0.1 m3 /s through the 100– mm–diameter nozzle and strikes the vane on the 150–kg cart, which is originally at res

t. Determine the velocity of the cart 3 seconds after the jet strikes the vane. Start your analysis by first properly selecting your C.V. and then present the corresponding FBD. rhow = 1000 kg/m3 .

Engineering
1 answer:
mamaluj [8]3 years ago
5 0

Answer:

10.19 m/s

Explanation:

See the attached picture.

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What are the parameters that affect life and drag forces on an aerofoil?
Vinil7 [7]

Answer:

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Explanation:

Drag force:

 Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.

Drag force given as

F_D=\dfrac{1}{2}\rho\ A\ V^2

So we can say that drag force depends on following properties

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

6 0
3 years ago
A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p
valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

a)

Cycle time

T_c=\dfrac{60E}{R_P}

T_c=\dfrac{60\times 0.95}{24}

Cycle time = 2.37 min

b)

Numbers of workers ,W

W=\dfrac{TWC}{T_c}

W=\dfrac{50}{2.37}

W= 21

Numbers of workers =21

c)

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time -  Time for repositioning

Ts = Tc- Tr

Ts= 2.37  - 9/ 60 min

Ts= 2.22 min

We  know that efficiency

\eta=\dfrac{TWC}{n.T_s}

0.94=\dfrac{50}{n\times 2.22}

n=23.94  ⇒n=24

n=24

Stations on the line =24

7 0
3 years ago
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton
SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}

where

k_{e} is constant

r is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

Thus F_{2}=126.955Newtons

5 0
3 years ago
The ______ number of a flow is defined as the ratio of the speed of flow to the speed of sound in the flowing fluid.
allsm [11]

Answer:

The statement is true

Explanation:

4 0
2 years ago
Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!
masha68 [24]
Of course they are needed because without them the society wouldn’t be as nice as it is right now and plus there would be no more buildings ! :)
8 0
3 years ago
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