Answer:
the energy absorbed is 4.477 x 10⁶ J
Explanation:
mass of the liquid, m = 13 kg
initial temperature of the liquid, t₁ = 18 ⁰C
final temperature of the liquid, t₂ = 100 ⁰C
specific heat capacity of water, c = 4,200 J/kg⁰C
The energy absorbed is calculated as;
H = mcΔt
H = mc(t₂ - t₁)
H = 13 x 4,200(100 - 18)
H = 4.477 x 10⁶ J
Therefore, the energy absorbed is 4.477 x 10⁶ J
The energy of the ski lift at the base is kinetic energy:
![K= \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20)
where m is the mass of the ski lift+the people carried, and
![v=15.5 m/s](https://tex.z-dn.net/?f=v%3D15.5%20m%2Fs)
is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
where
![g=9.81 m/s^2](https://tex.z-dn.net/?f=g%3D9.81%20m%2Fs%5E2)
and h is the height at the top of the hill.
So, since the total energy must conserve, we have
![U=K](https://tex.z-dn.net/?f=U%3DK)
and so
![mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20)
from which we find the height:
Answer:
3 m/s
Explanation:
<u>The velocity of a position-time graph is the slope of the line.</u> Slope is rise over run, or rise divided by run. The rise (how many units the line goes up) is 3 units and the run is 1 unit. 3/1 is 3, so the velocity is 3 m/s.
Answer:
![E=29\times 10^{-5}eV](https://tex.z-dn.net/?f=E%3D29%5Ctimes%2010%5E%7B-5%7DeV)
Explanation:
For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,
,
,
![E^{0}=E](https://tex.z-dn.net/?f=E%5E%7B0%7D%3DE)
Here, E is the energy in the absence of electric field.
And
are the highest and the lowest energies.
The difference of these energies
![\Delta E=2\mu B](https://tex.z-dn.net/?f=%5CDelta%20E%3D2%5Cmu%20B)
is known as Bohr's magneton.
B=2.5 T,
Therefore,
![\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J](https://tex.z-dn.net/?f=%5CDelta%20E%3D2%289.3%5Ctimes%2010%5E%7B-24%7DJ%2FT%29%5Ctimes%202.5%20T%5C%5C%5CDelta%20E%3D46.5%5Ctimes%2010%5E%7B-24%7DJ)
Now,
![Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV](https://tex.z-dn.net/?f=Delta%20E%3D46.5%5Ctimes%2010%5E%7B-24%7DJ%28%5Cfrac%7B1eV%7D%7B1.6%5Ctimes%2010%5E%7B-9%7DJ%20%7D%20%29%5C%5CDelta%20E%3D29.05%5Ctimes%2010%5E%7B-5%7DeV%5C%5CDelta%20E%5Csimeq29%5Ctimes%2010%5E%7B-5%7DeV)
Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is ![E=29\times 10^{-5}eV](https://tex.z-dn.net/?f=E%3D29%5Ctimes%2010%5E%7B-5%7DeV)