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Marina86 [1]
3 years ago
14

In the diagram, the liquid is vaporizing at which point?

Physics
1 answer:
sasho [114]3 years ago
3 0
E is the vapourising state
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Why can't we implement the center tapped full wave rectifier without center-tapped transformer
maksim [4K]
For a full wave bridge you don't want a center tap
6 0
3 years ago
Consider two tubes filled with water at the same height, one with fresh water and the other tube with salt water. The pressure i
Olegator [25]

Answer:

B

Explanation:

The correct option for the question is B that is salt water. In salt water, the density of water is higher so the pressure at the end of tube containing salt water will be greater. As according to the hydrostatic law the pressure at a given point will be directly proportional to the distance travelled as well.

3 0
3 years ago
Most of the mass of the Milky Way exists in the form of:<br> a.) starsb.) gasc.) dustd.) dark matter
Sveta_85 [38]
<h2>Answer: dark matter</h2>

It is believed that the Milky Way has 90% dark matter and only 10% ordinary matter (known matter). Because, like gravity, <u>dark matter can not be observed directly</u>, however its existence is inferred through the movement of the stars and the cosmic dust within the galaxy.

It is important to note that dark matter composition is unknown and corresponds to 80% of the matter in the universe. It does not emit or interact with any type of electromagnetic radiation, but it interacts with the known matter through gravity.  

5 0
3 years ago
4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx
frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

Ix=.\frac{Req}{Rx} *Itotal (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2}  } (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

Req.=\frac{ 100*R1}{R1+100} (3)

Now substituting (3) in (1).

25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0
3 years ago
Can a 20 N force and 40 N force ever produce a resultant with magnitude of 27 N?
SVEN [57.7K]

Sure ,Let's find angle between forces

  • Vectors be A and B and resultant be R

\\ \sf\longmapsto R^2=A^2+B^2+2ABcos\theta

\\ \sf\longmapsto 27^2=20^2+40^2+2(20)(40)cos\theta

\\ \sf\longmapsto 729=400+1600+1600cos\theta

\\ \sf\longmapsto 729=2000+1600cos\theta

\\ \sf\longmapsto 1600cos\theta=-271

\\ \sf\longmapsto cos\theta=-0.169

\\ \sf\longmapsto \theta=cos^{-1}(-0.169)

\\ \sf\longmapsto \theta=80.2°

7 0
2 years ago
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