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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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3 years ago
How is wavelength related to the different colors we see in a rainbow?
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3 years ago
A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be co
Naddika [18.5K]

Answer:

a = 45 m/s/s

Explanation:

As we know that total mass of the rocket is

M = 6000 kg

total mass of the fuel is given as

m = 2000 kg

all the fuel is burnt in 15 s

so rate of the fuel burning is given as

\frac{dm}{dt} = \frac{2000}{15}

\frac{dm}{dt} = 133.33 kg/s

now the thrust force on the rocket is given as

F_{th} = v\frac{dm}{dt}

(6000 - 133.33 t) a = (900 + v)(133.33)

so we have

\frac{dv}{900 + v} = (133.33)\frac{dt}{6000 - 133.33 t}

so we have

ln(\frac{900 + v}{900}) = - ln(\frac{6000 - 133.3 t}{6000})

1 + \frac{v}{900} = \frac{6000}{6000 - 133.3 t}

now acceleration is rate of change in velocity

\frac{1}{900}\frac{dv}{dt} = \frac{133.3\times 6000}{(6000 - 133.3t)^2}

so acceleration at t = 15 s

a = 45 m/s^2

3 0
4 years ago
Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. Wha
PSYCHO15rus [73]

Answer:

21 Hz, 43 Hz and 22 Hz

Explanation:

The computation of the beat frequencies that are generated by this discordant combination is as follows:

As we know that

beat frequencies = |f_1  - f_2|

So

For the first one

= |349 Hz - 370 Hz|

= 21 Hz

For the second one

= |349 Hz - 392 Hz|

= 43 Hz

And, for the third one

= |370 Hz - 392 Hz|

= 22 Hz

5 0
3 years ago
What is the log of 4311​
WINSTONCH [101]
The log of 4311 is 8.3689251747471
5 0
3 years ago
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