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3 years ago

Which statement best explains why a wet saw used to cut through a fine optical crystals is constantly lubricated with oil

1 answer:

5 0

The major reason why a wet saw that is used in cutting optical crystals is constantly lubricated with oil is largely due to friction, lubrication decreases friction and minimises the increase of internal energy.

Friction is basically the coming together of two parts that rubs against each other, it causes heat and lead to deterioration of surfaces, it can also lead to wear and tear.

Although, despite it down side, friction is desire-able in most applications

e.g the brake system of vehicles, and if not for friction, we humans cannot walk without falling

learn more:

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Inthisexperiment,aspringforcewasusedtokeep moving object travelingin a circularpath.The size of a spring force should be proport

poizon [28]

Answer:

By calculation, it can be shown that;

K = $\frac{F_{angular}}{2\times x\times c}$

Whereby for constant K, as ${F_{angular}}$ increases, x also increases.

Explanation:

The experiment set up consisted of the use of a spring force to maintain the object in circular path.

The energy in the spring is given by

$\frac{1}{2}\cdot k\cdot x^2$ .

Rotational kinetic energy = $\frac{1}{2}$ ·I·ω²

Inertia, I = $\frac{1}{2}$ ·m·r²

ω = $\frac{v}{r}$

Substituting gives

Rotational kinetic energy = $\frac{1}{2}$ ·

= $\frac{1}{4}$ ·m· v²

Equating both equations gives

K = $\frac{2\times m\times v^2}{4\times x^{2} }$ = $\frac{1\times m\times v^2}{2\times x^{2} }$

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

$\frac{ m\times v^2}{2\times x\times c\times r }$ = $\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}$

Since $\frac{v^{2} }{r}$ = angular acceleration, α, then

m× $\frac{v^{2} }{r}$ = Angular force

Therefore K = $\frac{F_{angular}}{2\times x\times c}$

Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.

3 0

3 years ago

Cuál es el significado del diálogo intercultural

bagirrra123 [75]

Answer:

El diálogo intercultural es un intercambio de opiniones abierto y respetuoso entre individuos y grupos pertenecientes a diferentes culturas que conduce a una comprensión más profunda de la percepción global del otro

Explanation:

“Intercultural dialogue is an open and respectful exchange of views between individuals and groups belonging to different cultures that leads to a deeper understanding of the other's global perception.”

5 0

3 years ago

How is energy measured?

stepladder [879]

Because a Btu is so small, energy is usually measured in millions of Btus. 1 Btu = the amount of energy required to increase the temperature of one pound of water (which is equivalent to one pint) by one degree Fahrenheit. This is roughly the heat produced from burning one match.

<em>https://www.ucsusa.org/clean_energy/our-energy-choices/how-is-energy-measured.html</em>

5 0

3 years ago

What is the equivalent resistance of the circuit?

Sonbull [250]

Answer:

80Ω

Explanation:

Resistance in a series circuit (when one resistor is after another in the same line) is calculated with the following equation

Total resistance = R1 + R2 + R3...Rn

Substitute in your values and solve for total resistance

Total resistance = 60 + 20 = 80Ω

8 0

3 years ago

Read 2 more answers

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

4 years ago