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2 years ago

How do different factors affect solubility? Check all of the boxes that apply.

Physics

2 answers:

Karo-lina-s [1.5K]2 years ago

4 0

Answer:

Explanation:

Solubility is affected by current concentration, mechanical disturbance, temperature and pressure. So the answers are:

- Increasing temperature decreases the solubility of gases.

- Stirring increases the solubility of solids.

- Increasing pressure increases the solubility of gases.

- Decreasing the amount of solvent decreases the solubility of solids, liquids, and gases.

Alika [10]2 years ago

3 0

Answer:

Explanation:

Increasing temperature decreases the solubility of gases : true

Stirring increases the solubility of solids.: true

Increasing pressure increases the solubility of liquids : false

Decreasing temperature decreases the solubility of solids.: true

Increasing pressure increases the solubility of gases: true

Decreasing the amount of solvent decreases the solubility of solids, liquids, and gases : true

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Explanation:

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2 years ago

A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the

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Answer:

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3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$