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iVinArrow [24]
3 years ago
5

5) In the last part of step 7 of the procedure, you measured the resistance of the flashlight when it had no current passing thr

ough it. The resistance of the flashlight is different, however, when current is passing through it. Explain how your measurement of the resistance of the variable resistor obtained in part 7 is a valid approximation of the resistance of the flashlight when it had current passing through it. Is the resistance higher when the flashlight is on or off
Physics
1 answer:
Andreyy893 years ago
6 0

Answer:

Following are the responses to this question:

Explanation:

The small current passes thru the capacitor of the strain gauge and the current is generated throughout the resistor. For the very first time,  in contrast to what we calculate, its resistance of the multimeter is quite high and therefore the small stream flowing through the bulb would have very little impact on the measure. Thus, as the current flows through the flashbulb, this same calculation is of excellent price, its material is heated and resistance varies with increase. Therefore, when the bulb will be on, sensitivity is greater.

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6 0
3 years ago
Read 2 more answers
Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconduc
julsineya [31]

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

8 0
3 years ago
Whats the answer???????????
Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

R_{1}=4\Omega

R_{2}=6\Omega

R_{3}=8\Omega

Now, when the resistors are connected in parallel, the total resistance R is calculated as follows:

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

Isolating R:

R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}

Rewriting with th known values:

R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}

Finally:

R=1.84 \Omega

Hence, the correct option is less than 4 ohms.

4 0
3 years ago
The heating of the filament is what causes the light production (photon emission), and heating is caused by the current in the l
alexira [117]

Answer:

explained

Explanation:

Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb

(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.

8 0
3 years ago
Part 1: Fill in the SI unit for each of the following measurements. 1. Time: 2. Length: 3. Mass: 4. Temperature:
Alik [6]

Answer:

1. Time: Second

2. Length: Meter

3. Mass: Kilogram

4. Temperatur: Kelvin

Explanation:

...

3 0
3 years ago
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