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Alex
3 years ago
7

1 gram of radium is reduced by 3.1 mg in 5 years by alpha decay.

Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

1 gm = 1000 mg

N = No e^-y t       where y = lambda the decay constant

N / N0 = .9969

ln .9969 = - y t

-.00310 = - 5 y

y = .00310 / 5 = .000621

So the half-life = .693 / y

T^1/2 = .693 / .000621 = 1116 or about 1120 yrs

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Why did she use an red infra lamp
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To act as the Sun' was accepted but if you put 'sunlight' alone it was not accepted. The examiner wanted you to state that the infra red radiation was needed to warm up the water.
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3 years ago
The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a
emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\  I_2= 0.125 \ foot-candles

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

#SPJ1

6 0
2 years ago
Which property of gold allows it to be used this way?
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the electric conductivity of gold is very high

3 0
4 years ago
how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy
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This question is probably referring to heat energy transferring from the car to its surroundings.
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3 years ago
You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

4 0
3 years ago
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