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3 years ago

Complete this sentence. The solubility of a sample will ____________ when the size of the sample increases.

Physics

1 answer:

Murljashka [212]3 years ago

6 0

The correct answer is B.

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3. How does Earth’s rotation affect our view of stars

Elis [28]

Since we ride along with the Earth while it's doing whatever it does,
the Earth's rotation causes our eyes to constantly point in a different
direction.

If we try to keep watching one star, we have to keep changing the
direction of our eyes to keep looking at the same star.

We can't feel the Earth rotating, so our brains say that the star ... and
the sun and the moon too ... is actually moving across the sky.

5 0

3 years ago

Read 2 more answers

When the cornea and lens of the eye do not properly focus

frutty [35]

Answer:

When there is nearsightedness or myopia

Explanation:

As, in myopia the image is formed in front of the retina.

Which makes things looking at things near very easy, but looking at far away things very difficult.

So concave lens makes things look bigger, so therefore, it is used during myopia, to make things look bigger when they are far away

3 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$