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2 years ago

A solution is oversaturated with solute. which could be done to decrease the oversaturation?

Physics

2 answers:

Rzqust [24]2 years ago

5 0

Answer: The correct awnser is B add more solute

Explanation:

Grace [21]2 years ago

4 0

<span>A solution is oversaturated with solute. The thing that could be done to decrease the oversaturation is to add more solvent in order to decrease the concentration of the solute. You can also increase the temperature to increase solubility of the solute. Hope this answers the question.</span>

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Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

$V = i R$

$i = \frac{V}{R}$

$i = \frac{120}{160}$

$i = \frac{3}{4} A$

now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

$Q = \frac{3}{4}(8 \times 60 \times 60)$

$Q = 21600 C$

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2 years ago

A gold puck has a mass of 12 kg and a velocity of 5i – 4j m/s prior to a collision with a stationary blue puck whose mass is 18

Ugo [173]

Answer:

Explanation:D

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2 years ago

A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

$p_{rad} = \frac{I}{c}$

<u>Where:</u>

$p_{rad}$ : is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

$p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}$

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

If you touch the two terminals of a power supply with your two fingertips on opposite hands, the potential difference will produ

LiRa [457]

Answer:

Yes the body will receive a dangerous shock in both cases.

Explanation:

Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.

Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.

Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:

I = V/R= 120/ 500*10^3

I= 0.24 mA

Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.

Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:

I = 120 / 1000

I = 12 mA

The current is higher than safe zone so the body will receive a dangerous shock.

7 0

2 years ago

A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.

kakasveta [241]

The first law of thermodynamics says that the variation of internal energy of a system is given by:
$\Delta U = Q + W$
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
$\Delta U = -1275 J+855 J=-420 J$

6 0

2 years ago