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andriy [413]
3 years ago
14

.Ryan boiled a liter of water and then stirred sugar into it, adding more sugar until no more would dissolve in the water, creat

ing a saturated solution. If he pours more sugar into it after it has had a chance to cool, what will most likely happen?Immersive Reader (1 Point) All of the sugar will come out of solution, and pure water will float to the top If he stirs constantly, the sugar will form into one large sugar crystal The added sugar will sink to the bottom The added sugar will dissolve in the water
Physics
1 answer:
kaheart [24]3 years ago
6 0

Answer: The correct option is that all of the sugar will come out of solution, and pure water will float to the top

Explanation:

Solution in the field of Chemistry is usually made up of two or more substances which contains a solute that dissolves in a solvent.

A solution can either be:

-> Saturated

--> Unsaturated or

-> Supersaturated.

A saturated solution is a solution with solutes that dissolves until it is unable to dissolve anymore leaving the undissolved solute beneath.

When there is mixture of a solute and a solvent in a solution the reactions that occurs are called crystallization and dissolution. Crystallization causes solid solutes to remain undissolved while dissolution is simply the dissolving process of the solute.

When Ryan added more sugar after reaching the saturation point of the mixture, the process of crystallization set in which surpassed the process of dissolution of the sugar solute leading to precipitation of the solute of out the solution.

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What is the acceleration of a car that goes from 40km/hr to 80km/hr in 2s?
tresset_1 [31]
Acceleration = v2 - v1 / t
a = 80 - 40 / 2
a = 40 / 2
a = 20 m/s²

In short, Your Answer would be 20 m/s²

Hope this helps!
5 0
3 years ago
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?
Orlov [11]
  • Force=10N
  • Displacement=30cm=0.3m

\\ \sf\bull\longmapsto Wd=Force(Displacement)

\\ \sf\bull\longmapsto Work=10(0.3)

\\ \sf\bull\longmapsto Work=3J

5 0
3 years ago
Read 2 more answers
In the experiment, “Rolling Along”, which ball had the greater mass?
olga2289 [7]

Answer:

b

Explanation:

4 0
3 years ago
You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an
Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0
3 years ago
On august 10, 1972, a large meteorite skipped across the atmosphere above the western united states and western canada, much lik
RUDIKE [14]
(a)
The velocity of the meteorite just before hitting the ground is:
v=20 km/s=20000 m/s
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
\Delta K =  \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J

(b) 1 megaton of tnt is equal to 1 MT=4.2 \cdot 10^{15}J
To find to how many megatons the meteorite energy loss \Delta E
corresponds, we can set the following proportion
1 MT: 4.2 \cdot 10^{15}J=x: \Delta E
And so we find
x=  \frac{\Delta E}{4.2 \cdot 10^{15}J}  = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of \Delta E=0.162 MT=162 kT. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
1:13 kT=x:162 kT
And so we find
x= \frac{162 kT}{13 kT}=12.46
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
7 0
4 years ago
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