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3 years ago

10

In a popular classroom demonstration, a cotton ball is placed in the bottom of a strong test tube. A plunger fits inside the tub

e and it makes an air-tight seal. It is then pushed down very rapidly, and the cotton flashes and burns. This happens because ______.

1 answer:

mario62 [17]3 years ago

3 0

Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.

$P = kT$

<em>where P: is the gas pressure, T: is the gas temperature and k: is a constant.</em>

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.

I hope it helps you!

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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

4 years ago

Two identical waves are destructively interfere. What will happen to the resulting wave?

storchak [24]

Answer:

Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference

Explanation:

8 0

3 years ago

Can the kinetic energy of an object be negative? Can the potential energy of an object be negative? Can a potential energy funct

IceJOKER [234]

Answer:

Kinetic energy cannot be negative

potential energy is a reference dependent quantity and it can be positive as well as negative both

Since potential energy is defined only for conservative force so it can not be found for friction force

Explanation:

Kinetic energy of an object is given by the formula

$KE = \frac{1}{2}mv^2$

here we know that

m = mass of object that can not be negative

v = speed of the object and since its square is given here so it can not be negative

so Kinetic energy is always positive

potential energy is given as the energy due to the virtue of the position of object

so it is

$\Delta U = -\int F.dr$

so potential energy is a reference dependent quantity and it can be positive as well as negative both

Since potential energy is defined only for conservative force so it can not be found for friction force

5 0

3 years ago

If a planet has a non-circular orbit around a star, as the planet moves closer towards a star its orbital speed ….

makkiz [27]

I think the answer you’re looking for is
D)
-I hope this helps! Enjoy the rest of your day

3 0

3 years ago

If the coefficient of static friction at d is μd = 0.3, determine the smallest vertical force that can be applied tangentially t

const2013 [10]

Answer:

The minimum force is required is, f = 0.3 mg Newton

Explanation:

Given data,

The coefficient of the static friction, μₓ = 0.3

Let the mass of the wheel is, m = M kg

The minimum force required to set the wheel in motion is equal to the force of static friction. This amount of force can be applied tangentially to the wheel in a vertical direction to set the wheel in motion.

<em> Fₓ = μₓ. F</em>

Where F is the normal force acting on the wheel and it is equal to mg.

Therefore, the minimum force is required is, f = 0.3 mg Newton

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4 years ago