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skelet666 [1.2K]
3 years ago
13

When measuring an unknown voltage with an analog VOM, you should first

Physics
2 answers:
Soloha48 [4]3 years ago
8 0
If you have no idea what the voltage is that you're about to measure,
then you should set the meter to the highest range before you connect
it to the two points in the circuit. 

Analog meters indicate the measurement by moving a physical needle
across a physical card with physical numbers printed on it.  If the unknown
voltage happens to be 100 times the full range to which the meter is set,
then the needle may find itself trying to move to a position that's 100 times
past the highest number on the meter's face.  You'll hear a soft 'twang',
followed by a louder 'CLICK'.  Then you'll wonder why the meter has no
needle on it, and then you'll walk over to the other side of the room and
pick up the needle off the floor, and then you'll probably put the needle
in your pocket.  That will end your voltage measurements for that day,
and certainly for that meter. 

Been there.
Done that.
weeeeeb [17]3 years ago
4 0

Answer:set the switch to the highest range

Explanation:

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A star produces 2x10^26 watts. how much energy does it lose every minutes
Step2247 [10]

Answer:

Energy loss per minute will be 120\times 10^{26}j

Explanation:

We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

We multiply with 60 we have to calculate energy loss per minute

7 0
4 years ago
The diagram does not represent a real electric field because the field lines, can someone help explain this for me
Papessa [141]

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible


4 0
3 years ago
A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w
Alecsey [184]

Answer:

581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

Q=+757 kJ

\Delta U = +176 kJ

Therefore the work done by the system is

W=Q-\Delta U=757 kJ-176 kJ=+581 kJ

And the positive sign means the work is done BY the system.

5 0
3 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
HELP ASAP PLEASE<br><br> Which of the following mechanical waves have the same energy?
serious [3.7K]

Answer:

1 and 4 is the answer

6 0
3 years ago
Read 2 more answers
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