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Temka [501]
4 years ago
5

13. A nuclear decay chain represents

Physics
1 answer:
murzikaleks [220]4 years ago
5 0

Answer:

a series of  decays producing sequentially more stable nuclei

Explanation:

A nuclear decay chain represents a series of decays producing sequentially more stable nuclei.

  • For every atomic nucleus, there is a specific neutron/proton ratio which ensures the stability of the nucleus.
  • Any nucleus with a neutron/proton combination different from its stability ratio (i.e either too many neutrons or too many protons) will be unstable and split into one or more other nuclei with the attendant emission of small particles of matter.
  • The decay process is a continuous chain reaction steps.
  • Until stability of the nucleus is attain, it continues.
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The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl
aleksandr82 [10.1K]
acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
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4 years ago
The major problem with hydrogen as an alternative source of energy is that none exists on the surface of the earth. True or fals
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3 years ago
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Sindrei [870]

Answer:

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Explanation:

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3 0
3 years ago
A student uses a stopwatch to measure the period of the pendulum of the Beverly clock in the corridor. His measurements are: (a)
Westkost [7]

Answer:

Reading is close to (b) 13.44 which is the best estimate of the period

Associated error, \Delta E =0.178 s

Given:

t_{a} = 13.54 s

t_{b} = 13.44 s

t_{c} = 13.89 s

t_{d} = 13.41 s

t_{e} = 13.17 s

t_{f} = 13.22 s

Solution:

1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:

Mean, \bar {x} = \fra{sum of all observations}{No. of observation}

Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}

Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}

Mean, \bar {x} = 13.445 s

It is close to 13.44 s

2. Associated error is given by:

\Delta E_{n} = |measured value - actual value|

\Delta E_{n} = |t_{n} - \bar {x}|

where

n = a, b,......, e

Now,

\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01

\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00

\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45

\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03

\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027

\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10

Mean Absolute Error, \Delta E = \frac{\Sigma E_{n}}{6}

\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}

\Delta E =0.178 s

3. The assumption behind the estimation is population is considered to distributed normally.

6 0
3 years ago
Two out of phase loudspeakers are some distance apart. A person stands 4.50 m from one speaker and 3.80 m from the other. What i
Contact [7]

Answer:

The third lowest frequency is 1489.27 Hz.

Explanation:

Given that,

Distance of person from one speaker= 4.50 m

Distance of person from the other = 3.80 m

The longest wavelength is the shortest frequency.

We need to calculate the difference between the speakers

The difference between the speakers d= 4.50-3.80 =0.7 m

We know that,

The longest wavelength is such that one wavelength

\lambda = d

So, The third longest wavelength is

3\lambda=d

Put the value of d

\lambda=\dfrac{0.7}{3}

\lambda=0.233\ m

We need to calculate the lowest frequency

Using formula of frequency

f=\dfrac{v}{\lambda}

f=\dfrac{347}{0.233}

f=1489.27\ Hz

Hence, The third lowest frequency is 1489.27 Hz.

4 0
3 years ago
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