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3 years ago

A student mixes two substances in a large beaker. After two hours, the student sees that the substances have separated again.

Physics

1 answer:

S_A_V [24]3 years ago

7 0

The full question is found in the image attached

Answer:

Densities

Explanation:

When we talk about density, one of the things that come to our mind is actually how heavy something is.

If i have substances that possess different densities, i will notice that i can not really mix them because they will separate and the heavier substance will be at the bottom while the lighter substance will be at the top.

Hence the two substances separated due to difference in density.

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A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti

melisa1 [442]

1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

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3 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

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3 years ago

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

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3 years ago