Answer:
<h2>
2113 seconds</h2>
Explanation:
The general decay equation is given as
, then;
where;
is the fraction of the radioactive substance present = 1/16
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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B16%7D%20%3D%20e%5E%7B-%5Clambda%288450%29%7D%20%20%5C%5C%5C%5CTaking%5C%20ln%5C%20of%20%5Cboth%20%5C%20%20sides%5C%5C%5C%5Cln%28%5Cfrac%7B1%7D%7B16%7D%20%29%20%3D%20%20ln%28e%5E%7B-%5Clambda%288450%29%7D%29%20%20%5C%5C%5C%5C%5C%5Cln%20%28%5Cfrac%7B1%7D%7B16%7D%20%29%20%20%3D%20-8450%20%5Clambda%5C%5C%5C%5C%5Clambda%20%3D%20%5Cfrac%7B-2.7726%7D%7B-8450%7D%5C%5C%20%5C%5C%5Clambda%20%3D%200.000328)
Half life f the material is expressed as ![t_{1/2} = \frac{0.693}{\lambda}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B0.693%7D%7B%5Clambda%7D)
![t_{1/2} = \frac{0.693}{0.000328}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B0.693%7D%7B0.000328%7D)
![t_{1/2} = 2,112.8 secs](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%202%2C112.8%20secs)
Hence, the half life of the material is approximately 2113 seconds
Answer:
True
Explanation:
As the Earth goes around the Sun, it will appear that Earth is stationary and Sun is going around it. One can observe the same in real life as well. This apparent path followed by the Sun is called Ecliptic. The plane consisting Ecliptic is called as Ecliptic plane which is same as the orbital plane of Earth.
All the planets of the Solar system are also going around the Sun. Their orbital plane has negligible tilt with respect to Ecliptic plane. Due to this the planets will always appear near to the Ecliptic as they move on the celestial sphere.
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
![m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}](https://tex.z-dn.net/?f=m_%7Bm%7D%20c_%7Bm%7D%20%28T_%7Bmi%7D%20-%20T%29%20%3D%20m_%7Ba%7D%20c_%7Ba%7D%20%28T%20-%20T_%7Bai%7D%29%20%2B%20m_%7Bw%7D%20c_%7Bw%7D%20%28T%20-%20T_%7Bwi%7D%20%29%20%5C%5C%28400%29%20%28100%20-%2040%29%20c_%7Bm%7D%20%3D%20%28100%29%20%28900%29%20%2840-%2015%29%20%2B%20%28500%29%20%284186%29%20%2840%20-%2015%29%5C%5C%20c_%7Bm%7D%20%3D%202274%20Jkg%5E%7B-1%7DK%5E%7B-1%7D)
Answer:
Explanation:
The direction of propagation of electromagnetic wave
is given by the direction of vector E x B where E is electrical field , B is magnetic field .
Given Electric field = E i because it is along x axis
Magnetic field = Bj because it is along y axis
E x B = Ei x Bj
= EB k .
so direction of E x B is along k direction or z - axis so wave is propagating along z - axis .