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dangina [55]
3 years ago
15

A burning candle is covered by a jar as shown in the picture. The whole arrangement has a mass of 500 g. What will be the approx

imate mass of the arrangement when the candle is completely burnt, after four minutes?
A)
0 g



B)
50 g



C)
250 g



D)
500 g
Physics
2 answers:
topjm [15]3 years ago
7 0

Technically, I can't answer the question, because you won't
let me see the picture that goes along with it and is a part of it. 
But I'm familiar with the set-up, have dealt with the question before,
and I can answer it from my previous experience and general knowledge.

If there is 500g of mass inside the jar when you lower it over
the candle, then there will be 500g of mass at any time after that,
forever, or until you pick up the jar and take some mass out or put
some more in.  It doesn't matter how long you wait.  It also doesn't
matter whether or not the candle is burning, whether or not the sun
is shining on the jar, or whether somebody comes along and spray-paints
the outside of the jar with black paint.  Matter is not created or destroyed. 
Whatever mass was inside when the jar got closed stays in there.
 
weqwewe [10]3 years ago
7 0
The answer is D)<span>500 g  it will not change</span>
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Frequency= speed/ wavelength
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=5.0 Hz

The answer would be letter D.
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In a circuit, energy is transferred to a charge.<br> Where is this energy transferred from?
vladimir1956 [14]

Answer:

Electric current.

Explanation:

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So it's the same with electrons.

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3 years ago
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

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L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

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8 0
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A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effect
Fantom [35]

Answer:

I_s=5.8A

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

P_p=P_s

So:

V_p*I_p=V_s*I_s

Where:

V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil

Solving for I_s

I_s=\frac{V_p*I_p}{V_s}

Replacing the data provided:

I_s=\frac{110*29}{550} =5.8A

4 0
3 years ago
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