Answer:
![I_2 = 3.16 I_1](https://tex.z-dn.net/?f=I_2%20%3D%203.16%20I_1)
Explanation:
given,
Intensity radiated by black body= I₁
room temperature = 300 K
Intensity radiated, I₂ = ?
At temperature = T₂ = 400 K
now,
Intensity of radiation, I = σ T⁴
now,
![\dfrac{I_1}{I_2}=\dfrac{T_1^4}{T_2^4}](https://tex.z-dn.net/?f=%5Cdfrac%7BI_1%7D%7BI_2%7D%3D%5Cdfrac%7BT_1%5E4%7D%7BT_2%5E4%7D)
![\dfrac{I_2}{I_1}=\dfrac{T_2^4}{T_1^4}](https://tex.z-dn.net/?f=%5Cdfrac%7BI_2%7D%7BI_1%7D%3D%5Cdfrac%7BT_2%5E4%7D%7BT_1%5E4%7D)
![I_2=I_1\dfrac{T_2^4}{T_1^4}](https://tex.z-dn.net/?f=I_2%3DI_1%5Cdfrac%7BT_2%5E4%7D%7BT_1%5E4%7D)
![I_2=I_1\dfrac{400^4}{300^4}](https://tex.z-dn.net/?f=I_2%3DI_1%5Cdfrac%7B400%5E4%7D%7B300%5E4%7D)
![I_2 = 3.16 I_1](https://tex.z-dn.net/?f=I_2%20%3D%203.16%20I_1)
hence, Radiation Intensity at 400 K is equal to I₂ = 3.16 I₁
Answer:
Explanation:
Speed of electron
v = 4 × 10^5 •j m/s
Magnetic field
B = 5 × 10^-5 T at angle of 45° to horizontal
Charge of electron
q = 1.6 × 10^-19C
Magnitude of force F?
The Force exerted in an electric field is given as
F = q(v×B)
Now, x component of the magnetic field
Bx = BCos45 = 5×10^-5 Cos45
Bx = 3.54 × 10^-5 •i T
Also, y component
By = BSin45 = 5 × 10^-5Sin45
By = 3.54 × 10^-5 •j T
B = 3.54 × 10^-5 •i + 3.54 × 10^-5 •j T
Now, F = q(v×B)
Note that,
i×i=j×j=k×k=0
i×j =k, j×k = i, k×i = j
j×i = -k, k×j = -i and i×k = -j
Therefore
F = q(v×B)
F = 1.6×10^-19(4×10^5•j × (3.54 × 10^-5 •i + 3.54 × 10^-5 •j T))
F = 1.6×10^-19 (4×10^5 × 3.54 × 10^-5 (j×i) + 4×10^5 × 3.54 × 10^-5(j×j))
F = 1.6×10^-19(14.14(-k) + 0)
F = —2.26 × 10^-18 •k N
It is in the negative direction of z axis
The magnitude of the force the field experience is 2.26 × 10^-18 N
Answer:
42 Hz
Explanation:
Given,
Speed ( v ) = 210 m/s
Wavelength ( λ ) = 5 m
To find : Frequency ( f ) = ?
Formula : -
v = f λ
f = v / λ
= 210 / 5
f = 42 Hz
Answer:
Energy transformation
Explanation:
Energy transformation is when energy changes from one form to another.