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Simora [160]
3 years ago
15

Calculate the activity of 60Co after 72hr, if you know the activity at time 0 is 35MBq (decay constant=0.001 day-1)?

Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

The activity of cobalt-60 after 72 hours is 34.895 MBq

Explanation:

A(t) = Ao(0.5)^t/t1/2

A(t) is the activity of cobalt-60 after time t

Ao is the initial activity of cobalt-60 = 35 MBq

t is time taken to reduce in activity = 72 hours = 72/24 = 3 days

t1/2 is the half-life = 0.693 ÷ decay constant = 0.693 ÷ 0.001/day = 693 days

A(72) = 35(0.5)^3/693 = 35 × 0.5^0.00433 = 35 × 0.997 = 34.895 MBq

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Likurg_2 [28]

Answer: During sexual reproduction, the genetic material of two individuals is combined to produce genetically-diverse offspring that differ from their parents.  The genetic diversity of sexually-produced offspring is thought to give species a better chance of surviving in an unpredictable or changing environment

7 0
3 years ago
What are the answers please help
aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.)  10.31 cm

e.)  10.24 cm  

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

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d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.)  If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm  

3 0
4 years ago
Why does wave height increase in shallow water??
Vitek1552 [10]
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3 0
3 years ago
Find the GCF 12, 18, and 84
ahrayia [7]
To find the answer, plot down the factors for every number.

12:   1, 2 ,3 ,4, 6, 12

18:   1, 2, 3, 6, 9, 18

84:  1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
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5 0
3 years ago
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Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:

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Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³

Thus, the total amount of bacteria is:

2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria
7 0
4 years ago
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