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sergiy2304 [10]
3 years ago
6

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

nute. At t = 4 hours, it measures 5 counts/minute. Fill in the table for the activity at the unknown times (in counts/minute).
Time Geiger Counter Radiation Rate
0 hours
1 hour 40 counts/minute
2 hours
3 hours
4 hours 5 counts/minute
Physics
1 answer:
Mama L [17]3 years ago
6 0

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

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Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

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This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

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v=u+at\\\Rightarrow a=\frac{v-u}{t}

From Newton's second law

F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}

Squarring both sides

112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m

The height from which the student fell is 5.7141 m

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3 years ago
A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction
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Answer:

The magnitude of the impulse delivered to the baseball is 7.0 Ns

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Impulse is given as change in momentum;

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Pi = 0.14 x 40 = 5.6 Ns

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Pf = 0.14 x 30

Pf = 4.2 Ns

The resultant impulse is given by;

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J² = 49

J = √49

J = 7.0 Ns

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Some physical properties of matter include:

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help.
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Answer:

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Explanation:

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v = Final velocity

s = Displacement

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The height of the cliff is 532.0725 m

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