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Hoochie [10]
3 years ago
11

15 points just because i’m feeling like super nice right now! brainly if you tell me how your day was? <3

Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:

it is good so far

i just passed my math test

Explanation:

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An electric circuit where electrons are free to flow
maw [93]

Answer: Electron flow / Electric current

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3 years ago
What mass of silver can be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s what mass of silver can
Schach [20]
To determine the mass plated, we use Faraday's Law of Electrolysis. We calculate as follows:

q = It
q = 8.70 (33.5) (60)
q = 17487 C

mass = 17487 C ( 1 mol e- / 96500 C) ( 1 mol / 2 mol e-) (107.9 g /mol)
mass = 9.78 g

Hope this helps.

4 0
3 years ago
Read 2 more answers
15. What is one way in which scientists study dark
Kay [80]

Answer:

It's d

Explanation:

Scientists study dark matter by looking at how it affects visible matter.

8 0
3 years ago
The gage pressure in a liquid at a depth of 3 m is read to be 50 kPa. Determine the gage pressure in the same liquid at a depth
Andrei [34K]

Answer:

The Gauge pressure at 9 meters depth is 150 \, kPa

Explanation:

Gauge pressure is the difference between absolute pressure and some reference pressure, most commonly atmospheric pressure. The increment in pressure caused by a static fluid is given by:

\Delta P = \rho g d where \rho is the density of the liquid,  g is the accleration due to gravity and d is the depth.

Now, we see that \Delta P is linearly proportional to d, and we can assume that \rho remains constant, because liquids are usually not compressible.

Given that the greater depth is simply 3 times the smaller depth:

d_2=3\cdot d_1\\9\,m= 3 \cdot 3\,m

\Delta P at 9\, m of depth will also be three times the gauge pressure at 3 \,m of depth.

We could also have calculated \rho ny using:

\Delta P = \rho \,g \,d\\\\\rho = \frac{\Delta P}{g \, d}\\\\\rho = \frac{\Delta P}{g \, d}= \frac{30 \,kPa}{9.8 \frac{m}{s^2}  \, 3\,m}=1020.41 \frac{kg}{m^3}

and used this result to calculate the gauge pressure. These are both similar methods that yield the same result

4 0
3 years ago
Please help
adelina 88 [10]

Answer:

I'm pretty sure that the answer is A

5 0
3 years ago
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