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dexar
10 days ago
9

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.​

Physics
1 answer:
Yuliya22 [2.9K]10 days ago
6 0
Factors influencing friction

The magnitude of friction is contingent on the following elements: i) The surface area in contact. ii) The applied pressure on the surfaces. Force is determined by Pressure multiplied by Area; thus, if the contact area increases or if the pressure applied rises, the frictional force will also escalate.

Methods for reducing friction

i) Smooth the contact surface. ii) Apply oil or grease to fill small gaps in flat surfaces. iii) Use ball bearings to minimize contact area among rotating components.

Lubrication

To minimize friction, various methods may be employed: Oil can be either thin or viscous, which depends on its SAE number (SAE indicating Society of Automotive Engineers). Highly viscous oils may not reach all components effectively. In contrast, very thin oils may drain away quickly, resulting in wastage. Grease is preferable in such situations, particularly around ball-bearings. Regular grease or oil should not be utilized under high speed, high pressure, and high temperature conditions—specialized lubricants are required then. The consistency of oil varies with temperature; it thickens in the cold and thins in the heat. Therefore, the choice of lubricant should be seasonally appropriate, and it's always wise to consult the equipment's operating manual prior to making a selection.[[TAG_11]]
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The intensity of the sound increases because sound waves are mechanical waves, meaning they cannot move through a vacuum and require a medium to propagate.
4 0
15 days ago
An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Maru [2979]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

The worker is positioned at 2.5 m

7 0
28 days ago
A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
kicyunya [2911]

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

Now, the potential for the smaller droplets is given as:

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

7 0
27 days ago
Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin
Ostrovityanka [2827]
The equations are h = v₀² / 2g, and h = k/4g x².
8 0
11 days ago
Read 2 more answers
A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o
Ostrovityanka [2827]

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

   Em₀ = U = mg y

Bottom

    Em_{f} = K = ½ m v²

    Emo =Em_{f}

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

  v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

     F = ma

     a = v² / r

For the turning radius, the cable length is r = L.

    F = m 2g (1 - cos θ)

Now, let's find the result.

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N

   

7 0
14 days ago
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