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2 years ago

10

Which best describes solubility? the speed at which a substance dissolves the ability of one substance to dissolve in another th

e amount of surface area per a given mass the temperature at which subtances become a mixture

1 answer:

ycow [4]2 years ago

4 0

<u><em>Answer:</em></u>

The ability of a substance to dissolve into another, called the solvent.

<u><em>Explanation:</em></u>

That is why water is called <u><em>"the universal solvent."</em></u>

Because it can dissolve almost anything.

The speed and temperature have nothing to do with solubility.

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If the initial upward speed of the ball in Activity 7 C.2 is 10m/s, and the ball is releasedat a height of 1.5 m above the oor,

Ulleksa [173]

Answer:

6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.

Explanation:

Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches maximum level, $v_0$ = 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and $\Delta s$ is the distance traveled, which we care looking for:

$0^2 - 10^2 = -2*(-10)\Delta s$

$\Delta s = 100 / (2 * 10) = 5 m$

So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.

5 0

3 years ago

Read 2 more answers

The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i

Butoxors [25]

Answer:

$p(t)=\frac{1}{20}(1+2e^{-t/90})$

Explanation:

Let $p(t)$ be % of $CO_2$ at time $t$ (time in minutes).

Amount of $CO_2$ in room=Volume of room $\times p$

Rate of change of $CO_2$ in room=Volume of room x $\frac{dp}{dt}$

$=180\times \frac{dp}{dt}$

Rate of inflow of $CO_2$ = $(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Inflow\ of \ air)$

$=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}$

Rate of $CO_2$ outflow

$(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}$

Therefore rate of change of $CO_2$ is $\frac{1}{100}-\frac{p}{50}$

% rate of change is $100\times \ Rate of \ Change=\frac{1}{10}-2p$

Therefore we have:

$180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}$

Integrate both:

$\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}$

$\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C$

In $|20p-1|=-\frac{t}{90}+$ In $K$

$+20In \ C=+In \ K$

Raise both sides to base $e$ ,

$20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})$

Initially, there's only 0.25% of $CO_2$ ,

Now substitute $p=0.25$ and $t=0$

$0.25=\frac{1}{20}(1+K)\\K=4$

$p=\frac{1}{20}(1+2e^{-t/90})$

8 0

3 years ago

A toy car accelerates at 3 m/s2 for 2 seconds. Its final velocity is 15 m/s, What is its initial velocity?

Talja [164]

Answer:

can i have the toy car after your done asking questions with it

Explanation:

4 0

3 years ago

Naily [24]

Features of the mobilization monkey

4 0

3 years ago

Read 2 more answers

A basketball with a mass of 0.61 kg falls vertically to the floor where it hits with a velocity of −7 m/s. (We take the positive

fredd [130]

Answer:

a. What impulse act on the ball during its collision with the floor?

b. If the ball is in contact with the floor for 0.04 s, what is the average force of the ball on he floor?

The answers to the question are

a. The impulse acting on the ball during its collision with the floor is

7.076 kg·m/s

b. The average force of the ball on the floor with contact of 0.04 s is

176.9 N

Explanation:

Mass of the ball = 0.61 kg

Initial velocity = -7 m/s

Final velocity = 4.6 m/s

Impulse = Δp = FΔt = mΔv

Where Δt =change in time

m = mass of Average force of he the ball

Δv = velocity change

Therefore impulse = m×(Final velocity - initial velocity)

= 0.61 × (4.6-(-7)) = 0.61×11.6

= 7.076 kg·m/s

b. Average force of the ball on the floor is given by

$F_{Average}$ ×Δt = mΔv = 7.076 kg·m/s

where Δt = change in time = 4.6 m/s

Therefore $F_{Average}$ =mΔv /Δt = 7.076/0.04 = 176.9 kg·m/s²

= 176.9 N

6 0

4 years ago