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denpristay [2]
1 year ago
5

A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl

y above the power source is a dot labeled X and then a circle with an X in it. The circuit then splits with one path straight back to the power source and the other path has 3 circles with X in them labeled 2, 3, and 4 respectively. There is a point labeled Y between circles 2 and 3.A diagram of a circuit with a power source on the left. Directly above the power source is a dot labeled X and then a circle with an X in it. After the x are 4 different circles on the circuit with Xs in them labeled 1, 2, 3, and 4 respectively. There is a point labeled Y in between circles 2 and 3. There is a branch of the circuit from X to Y. How does the circuit change when the wire is added? A closed circuit occurs and makes all bulbs turn off. An open circuit occurs and makes all bulbs turn off. A short circuit occurs and makes bulbs 3 and 4 turn off but keeps bulbs 1 and 2 lit. A short circuit occurs and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit.
Physics
1 answer:
tino4ka555 [31]1 year ago
7 0

Answer:

c thats the answer yep

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1) the strength of an electromagnet can be increased by
miss Akunina [59]
Using coils of fewer turns on the electromagnet
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3 years ago
How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/
Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J

Therefore, amount of heat lost in this process is 16.736 J.

4 0
3 years ago
A submarine can detect an approaching enemy submarine using sonar. If a stationary submerged submarine emits a 100kHz tone and r
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Answer:

B) 1.5 m/s

Explanation:

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f = F \frac{V+v}{V-v}\\

\frac{f}{F} =\frac{V+v}{V-v}\\  

\frac{f}{F}-1  =\frac{V+v}{V-v}-1\\

\Delta f = \frac{2vf}{V-v}\\

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Answer:

Assuming you mean how they work,

Explanation:

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