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1 year ago

7

If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s

econd year?

1 answer:

Tasya [4]1 year ago

4 0

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2 * 1/2 = 1/4 <======= 1/4 would be left

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Disagree.
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What is the wavelength of a 100 mhz radio signal?

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3 years ago

If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of

Vinvika [58]

We will have the following:

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1 year ago

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

Tamiku [17]

Answer:

7 m .

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For destructive interference

Path difference = odd multiple of λ /2

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If I am 1 m away from B , the path difference will be

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So path difference becomes odd multiple of λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

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4 years ago

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At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi

Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and $\mu$ be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

v = 0, s = 84 m, $\mu$ = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

F = ma

$-\mu N$ = ma

$-\mu mg$ = ma

a = $-\mu g$

Also,

$v^{2} = u^{2} + 2as$

$(0)^{2} = u^{2} + 2(-\mu g)s$

$u^{2} = 2(\mu g)s$

= $\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}$

= 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

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3 years ago