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ivanzaharov [21]
3 years ago
7

As part of a chemistry experiment, Austin ends up with a beaker containing a mixture of two liquids, which he knows have boiling

points of 77 °C and 103 °C. Austin is already familiar with___________ as a good method for separating liquids, so he asks his chemistry teacher where to find the heating mantle and condensing tube he'll need to___________ the mixture.
Engineering
1 answer:
asambeis [7]3 years ago
7 0

Answer:

Distillation, heat

Explanation:

Here in this question, we simply want to look at the best options that could fit in the gaps.

We have a mixture of liquids having boiling points which is far from each other.

Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.

Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.

Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.

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In an oscillating LC circuit, L ! 25.0 mH and C ! 7.80 mF. At time t 0 the current is 9.20 mA, the charge on the capacitor is 3.
topjm [15]

Answer:

a) 926 μJ

b) 3.802 mC

c) 8.61 A

d) 0.0721

e) 3.2137

Explanation:

The energy in the inductor is

El = \frac{1}{2}*L*I^2

El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J

The energy store in a capacitor is

Ec = \frac{1}{2}*C*V^2

The voltage in a capacitor is

V = Q/C

V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V

Therefore:

Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J

The total energy is Et = 925.6 + 1.1 = 926.7 μJ

At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge

Ec = \frac{1}{2}*C*V^2

V = Q/C

Ec = \frac{1}{2}*C*(\frac{Q}{C})^2

Ec = \frac{1}{2}*\frac{Q^2}{C}

Q^2 = 2*Ec*C

Q = \sqrt{2*Ec*C}

Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC

When the capacitor is completely empty all the energy will be in the inductor and current will be maximum

El = \frac{1}{2}*L*I^2

I^2 = 2*\frac{El}{L}

I = \sqrt{2*\frac{El}{L}}

I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A

At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC

q = Q * cos(vt + f)

q(0) = Q * cos(v*0 + f)

3.8 = 3.81 * cos(f)

cos(f) = 3.8/3.81

f = arccos(3.8/3.81) = 0.0721

If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137

4 0
3 years ago
What does the word “robot” mean? A.Clone B. Athlete C. Servant D. Actor
hram777 [196]

Answer:

a. clone

Explanation:

4 0
3 years ago
Ultra-thin semiconductor materials are of interest for future nanometer-scale transistors, but can present undesirably high resi
Mademuasel [1]

Answer:

p = 2*10^(-7) ohm m

Explanation:

The resistivity and Resistance relationship is:

p = \frac{R*A}{L}

For lowest resistivity with R < 100 ohms.

We need to consider the possibility of current flowing across minimum Area and maximum Length.

So,

Amin = 2nm x 10 nm = 2 * 10^(-16) m^2

Lmax = 100nm

Using above relationship compute resistivity p:

 p = \frac{100*2*10^(-16)}{100*10^(-9)} \\\\p = 2 * 10^(-7)

Answer: p = 2*10^(-7) ohm m

6 0
3 years ago
How did technology change society in the Renaissance?
WINSTONCH [101]
I think it’s A


Hope I helped
8 0
3 years ago
Read 2 more answers
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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