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zhuklara [117]
3 years ago
11

The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.

Physics
1 answer:
Rainbow [258]3 years ago
8 0

Answer: 0.0353\ s^{-1}

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in 2.59\times 10\ s

Sample at any time is given by

N=N_oe^{-\lambda t}

where, \lambda=\text{decay constant}

Put values

\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10

Taking natural logarithm both side

\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}

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Examples of increase in pressure due to increase in applied force<br>​
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Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
Talja [164]
The solution to the problem is as follows:

 <span>Average = 80 
So Sum = 80 * 5 = 400 
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88). 
Three results are 81, 88, and 88. 
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62. 

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? 
d = 144 - 16t2 
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3 0
3 years ago
The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

8 0
3 years ago
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