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seropon [69]
3 years ago
9

At a sudden contraction in a pipe the diameter changes fromD1toD2. The pressure drop,Δp, which develops across the contraction i

s a function ofD1andD2, as well as the velocity,V, in the larger pipe, and the fluid density, rho, and viscosity,μ. UseD1,V, andμas repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer: The working and answer can be viewed from the screenshots below. Thanks

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A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache
Paladinen [302]

Answer:

F_a=1470\ N

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

\displaystyle F_{r2}-T=0

The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

8 0
2 years ago
Radiation measured in emissions/sec is called the curie. <br> a. True<br> b. False
iragen [17]
True is the correct anwser


3 0
3 years ago
Any two application of gravity
Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

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3 years ago
If two stars are the same size and one is twice the temperature of the other, how much more luminous is the hotter one? quizlit
Dmitriy789 [7]
The hotter star will be 16 times more luminous  - luminosity depends on two things  - the size of the star and the temperature of the star. The hotter a star is, the more energy it will give out. This will give rise to greater luminosity.
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The frequency of a wave is 560 Hz. What is it’s period
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The answer would be, "1/560 seconds".
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