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zloy xaker [14]
3 years ago
7

Americium-241 is used in smoke detectors. It undergoes alpha decay with a half life of 432 years. The alpha particles ionize som

e of the air near the americium-241 source, and these charged ions are collected on a metal plate that has a small voltage applied to it. The collected particles cause a small electric current to flow in the detector. If smoke particles get in between the americium-241 source and the detector and reduce the current, the alarm is triggered. A typical smoke detector experiences 370000 alpha decays each second. How much time is needed for the number of decays to reduce to 93000 decays per second?
Physics
1 answer:
babymother [125]3 years ago
3 0

Answer:

860.6 years.

Explanation:

The parameters given are;

Initial detector activity = 370000 alpha decays per second

Final detector activity = 93000 alpha decays per second

Formula for time to change in activity is given by the following relation;

t_{93000} = \dfrac{-ln\dfrac{A}{A_0} }{\lambda} =  \dfrac{-ln\dfrac{93000}{370000} }{5.08 \times 10^{-11}} = 2.72 \times 10^{10} \, seconds

t₉₃₀₀₀ = 2.72 × 10¹⁰ seconds = 860.6 years.

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A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
When is the velocity of a mass on a spring at its maximum value?
ehidna [41]

Answer:

A.  when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


8 0
3 years ago
Read 2 more answers
The SAF operates the M113 Ultra APC.
HACTEHA [7]

The volume of the object must be no larger than 11.15 m^3.

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

\rho = 1000 kg/m^3

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

\rho = \frac{m}{V}

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

m=1.115\cdot 10^4 kg

Therefore, we can re-arrange the equation to find its volume:

V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3

So, the volume of the object must be no larger than 11.15 m^3.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

8 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid
IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x x velocity^{2}

  • incoming kinetic energy = 0.5 x m x x u^{2}

     

  •  coefficient of restitution =\frac{v}{u}

       0.72 =\frac{v}{u}

       v = 0.72u

        therefore the outgoing kinetic energy = 0.5 x m x (0.72u)^{2}

        outgoing kinetic energy = 0.5 x m x 0.5184 x u^{2}

        outgoing kinetic energy = 0.5184 (0.5 x m x x u^{2})

recall that 0.5 x m x x u^{2} is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

5 0
3 years ago
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