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stealth61 [152]
3 years ago
5

Francium-223 has a half-life of 22 minutes. A Geiger counter gives a reading of 58 counts per minute (cpm) from a sample of fran

cium-223. What will the reading be 22 minutes later? You can ignore background radiation.
Physics
1 answer:
polet [3.4K]3 years ago
8 0

Answer:

29 counts per minute (29 cpm)

Explanation:

The half-life of a radioactive isotope is the time needed for the activity of the sample to halve.The half-life of Francium-223 is exactly 22 minutes: this means that its activity after 22 minutes becomes exactly half of its initial value. Since the initial activity of this sample of francium-223 was 58 cpm, the activity after 22 minutes will be

A=\frac{58 cpm}{2}=29 cpm

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Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
Justin notices a particular type of caterpillar feeds only on cottonwood trees in his neighborhood.In which way has Justin incre
Anna007 [38]
Justin notices a particular type of caterpillar feeds only on cottonwood trees in his neighborhood.In which way has Justin increased his scientific knowledge, Justin increased his powers of scientific observation. His observation is empirical evidence. Empirical relies on observation  and experimentation.

It will help you..........

7 0
2 years ago
Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone
anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0
3 years ago
What are the frequencies of EM spectrum
kap26 [50]

The EM spectrum has no limits.  Any frequency you can imagine
is the frequency of some electromagnetic radiation somewhere.


3 0
3 years ago
When a wave enters a new medium from an angle, both the speed and the ________ change
slamgirl [31]

Answer:

B: Amplitude

Explanation:

When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.

The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.

8 0
2 years ago
Read 2 more answers
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