Answer:
1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K
Explanation:
The relationship between temperature and thermal energy is
E = K T
The relationship of the speed of light
c =λ f = f / ν 1/λ= ν
The Planck equation is
E = h f
Let's start the transformations
c = f λ = f / ν
f = c ν
E = h f
E = h c ν
E = KT
h c ν = K T
T = h c ν / K =( h c / K) ν
Let's replace the constants
h = 6.63 10⁻³⁴ J s
c = 3 10⁸ m / s
K = 1.38 10⁻²³ J / K
v = 1 cm-1 (100 cm / 1 m) = 10² m-1
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²
A = h c / K = 1,441 10⁻²
T = 1.44K
ν = 103 cm⁻¹ = 103 10² m
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²
T = 148K
1 Rydberg = 1.097 10 7 m
As we saw at the beginning the λ=1 / v
T = (h c / K) 1 /λ
T = 1,441 10⁻² 1 / 1,097 10⁷
T = 1.3 10⁻⁹ K
E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J
E = KT
T = E/K
T = 1.6 10⁻¹⁹ /1.38 10⁻²³
T = 1.16 10⁴ K
Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
Answer:
Explanation:
The change in kinetic energy will be simply the difference between the final and initial kinetic energies: 
We know that the formula for the kinetic energy for an object is:
where <em>m </em>is the mass of the object and <em>v</em> its velocity.
For our case then we have:
Which for our values is:
The answer is 5. To find the advantage you just divide 20 by 4.
<h2>
Answer:</h2>
In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:
So the average power is:
But:
So:
![P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%28%5Cfrac%7B1%2Bcos%282%5Comega%20t%29%7D%7B2%7D%20%29dt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7Bcos%282%5Comega%20t%29%7D%7B2%7D%5Ddt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B4%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2T%7D%5B%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B2%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2%7D)
In terms of RMS values:
