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Rufina [12.5K]
3 years ago
6

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

s) of the material
Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as N = N_0e^{-\lambda t} \\\\, then;

\dfrac{N}{N_0} = e^{-\lambda t} \\ where;

N/N_0 is the fraction of the radioactive substance present = 1/16

\lambda is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

\frac{1}{16} = e^{-\lambda(8450)}  \\\\Taking\ ln\ of \both \  sides\\\\ln(\frac{1}{16} ) =  ln(e^{-\lambda(8450)})  \\\\\\ln (\frac{1}{16} )  = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328

Half life f the material is expressed as t_{1/2} = \frac{0.693}{\lambda}

t_{1/2} = \frac{0.693}{0.000328}

t_{1/2} = 2,112.8 secs

Hence, the half life of the material is approximately 2113 seconds

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Explanation:

Given data

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solution

we know

v = angular frequency / wave number

and

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angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

so

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when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2  m

and when x =  x = 1.59 m and t = 0.150 s

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y(x,t)=9.00 × 10^−2 × (0.99959)

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5 0
3 years ago
Length of table is 1.0m,1.00m and 1.000m.Which one is more accurate?​
natita [175]

Answer:

1.00 m is a more accurate measured length.

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Case 3: For L=1.000 m, there is three significant digits after the decimal.

When one more significant digit after decimal considered, the exact number can be from 0.0095 to 1.0005.

So, the possible span of error \Delta E_3= 1.0005-0.0095= 0.001m

As \Delta E_1 >\Delta E_2>\Delta E_3

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2 years ago
A cubical furnace with external dimensions of 0:6 m is constructed from reclay brick. If the wall thickness is 40 mm, the inner
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Answer:

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8 0
3 years ago
Please have a look at the image!
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3 0
3 years ago
A light rope is attached to a block with mass 4.10 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
musickatia [10]

Answer:

a)  please find the attachment

(b) 3.65 m/s^2

c) 2.5 kg

d) 0.617 W

T<weight of the hanging block

Explanation:

a) please find the attachment

(b) Let +x be to the right and +y be upward.

The magnitude of acceleration is the same for the two blocks.  

In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:  

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c) <em>in order to calculate m we will apply newton second law on the hanging   </em>

<em>    block</em>

<em> </em>∑F=ma_y

T-W= -ma_y

T-mg= -ma_y

T=mg-ma_y

T=m(g-a_y)

a_x=a_y

14.7=m(9.8-3.65)

 m = 2.5 kg

<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>

d) calculate the weight of the hanging block :

W=mg

W=2.5*9.8

  =25 N

T=14.7/25

 =0.617 W

T<weight of the hanging block

6 0
3 years ago
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