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kondor19780726 [428]
4 years ago
8

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

of firecracker A. You see two flashes of light, from the two explosions, at exactly the same instant of time. Define event 1 to be firecracker A explodes and event 2 to be firecracker B explodes. According to your lab partner, based on measurements he or she makes, does event 1 occur before, after, or at the same time as event 2? Explain
Physics
1 answer:
blsea [12.9K]4 years ago
3 0

Answer:

Explanation:

A and B are 600 m apart . firecrackers explodes and make two waves , sound waves and light waves . they spread  out on all sides . The person sitting in the middle , observes two lights simultaneously and after some time hears two sound simultaneously . It is so because waves coming from A and B covers equal distance to reach the middle point . Same will be the case for sound waves .

But for observer sitting on the other side of A , both light and sound wave will not reach simultaneously . Light wave coming from A will reach earlier because distance covered by light will be less . Similarly sound coming from A will reach earlier . So he / she will observe cracker at A to have exploded earlier than that at B .

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Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

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First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

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Answer:

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Explanation:

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