Answer:
Angle of refraction is 23° .
Explanation:
![\frac{ \sin(0w) }{ \sin(0a) } = \frac{n \: a}{n \: w}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csin%280w%29%20%7D%7B%20%5Csin%280a%29%20%7D%20%20%3D%20%20%5Cfrac%7Bn%20%5C%3A%20a%7D%7Bn%20%5C%3A%20w%7D%20)
![\frac{ \sin(30) }{ \sin \: 0a } = \frac{1.3}{1.0}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csin%2830%29%20%7D%7B%20%5Csin%20%5C%3A%200a%20%7D%20%20%3D%20%20%5Cfrac%7B1.3%7D%7B1.0%7D%20)
![\sin \: 0w \: = \: \frac{ \sin(30) }{1.3}](https://tex.z-dn.net/?f=%20%5Csin%20%5C%3A%200w%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%5Cfrac%7B%20%5Csin%2830%29%20%7D%7B1.3%7D%20)
![= 0.3846153846](https://tex.z-dn.net/?f=%20%3D%200.3846153846)
![\sin^{ - 1} (0.3846153846](https://tex.z-dn.net/?f=%20%5Csin%5E%7B%20-%201%7D%20%280.3846153846)
![0w \: = \: 23 ^{o}](https://tex.z-dn.net/?f=0w%20%5C%3A%20%20%3D%20%20%5C%3A%2023%20%5E%7Bo%7D%20)
Answer:
The electric flux though each face of the cube is 0.0942 Nm²/C.
Explanation:
The expression for the electric flux through the cuboid is given by:
![\phi_E=E\times A=\frac {q}{\epsilon_0}](https://tex.z-dn.net/?f=%5Cphi_E%3DE%5Ctimes%20A%3D%5Cfrac%20%7Bq%7D%7B%5Cepsilon_0%7D)
Since, area of cuboid = 6a²
![\phi_E=E\times 6a^2=\frac {q}{\epsilon_0}](https://tex.z-dn.net/?f=%5Cphi_E%3DE%5Ctimes%206a%5E2%3D%5Cfrac%20%7Bq%7D%7B%5Cepsilon_0%7D)
Where,
E is the electric field
a is the side of the cuboid
q is the charge
is the constant having value 8.85×10⁻¹² C²/Nm²
Thus, the expression for the electric flux through one face of the cuboid is given by:
![\phi_E=E\times a^2=\frac {q}{6\times \epsilon_0}](https://tex.z-dn.net/?f=%5Cphi_E%3DE%5Ctimes%20a%5E2%3D%5Cfrac%20%7Bq%7D%7B6%5Ctimes%20%5Cepsilon_0%7D)
So,
Given ,
q = 5.0×10⁻¹² C
![\phi_E=\frac {5\times 10^{-12}\ C}{6\times 8.85\times 10^{-12}\ C^2N^{-1}m^{-2}}](https://tex.z-dn.net/?f=%5Cphi_E%3D%5Cfrac%20%7B5%5Ctimes%2010%5E%7B-12%7D%5C%20C%7D%7B6%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5C%20C%5E2N%5E%7B-1%7Dm%5E%7B-2%7D%7D)
Solving,
<u>The electric flux though each face of the cube is 0.0942 Nm²/C.</u>
<u></u>
Transverse Waves: Displacement of the medium is perpendicular to the direction of propagation of the wave.
Longitudinal Waves: Displacement of the medium is parallel to the direction of propagation of the wave.
KE = 60,000 J
Explanation:
KE = (1/2)mv^2
= (1/2)(30000 kg)(2 m/s)^2
= 60,000 J
Answer:
Heat required to raise the temperature of aluminium is 2250 J
Explanation:
Heat required to raise the temperature of the given substance is
![Q = ms\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20ms%5CDelta%20T)
here we know that
m = 0.10 kg
specific heat capacity of the metal is
s = 900 J/kg C
change in temperature of the metal is 25 degree
Now we have
![Q = 0.10(900)(25)](https://tex.z-dn.net/?f=Q%20%3D%200.10%28900%29%2825%29)
![Q = 2250 J](https://tex.z-dn.net/?f=Q%20%3D%202250%20J)