Answer:
C
Explanation:
it transforms electrical energy into mechanical energy.
Answer:
hence option A is correct
Explanation:
heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature
heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ
( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100
J/(kg °C )
heat required from 0°C ice to 0°C water = mass× specific heat of fusion of ice
= 7×3.36×10^5
= 2.352 × 10^6 J
= 2.352 MJ
TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ
= 2.4843 MJ
hence option A is correct
4 N/cm^2 is the answer. the way you get this is the formula for pressure is P=F/A. So you would plug in the numbers which would make the equation P=12/3 which you know equals 4. The units for pressure is N/cm^2 or N/m^2
I hope this helps
Answer:a. Magnetic dipole moment is 0.3412Am²
b. Torque is zero(0)N.m
Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A
That is,
U = n*I*A
But Area A is given as pi*radius² since it is a circular coil
Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is
A = 3.142*(0.05)² =7.86*EXP {-3} m²
Current I is 2 A
Number of turns is 20
So magnetic dipole moment U is
U = 20*2*7.86*EXP {-3}=0.3142A.m²
b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U
Torque = B x U =B*U*Sine(theta)
But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m
Explanation:
Given that,
Wavelength = 6.0 nm
de Broglie wavelength = 6.0 nm
(a). We need to calculate the energy of photon
Using formula of energy
![E = \dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
![E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B6.63%5Ctimes10%5E%7B-34%7D%5Ctimes3%5Ctimes10%5E%7B8%7D%7D%7B6.0%5Ctimes10%5E%7B-9%7D%7D)
![E=3.315\times10^{-17}\ J](https://tex.z-dn.net/?f=E%3D3.315%5Ctimes10%5E%7B-17%7D%5C%20J)
(b). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
![\lambda=\dfrac{h}{\sqrt{2mE}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2mE%7D%7D)
![E=\dfrac{h^2}{2m\lambda^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bh%5E2%7D%7B2m%5Clambda%5E2%7D)
Put the value into the formula
![E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B%286.63%5Ctimes10%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes9.1%5Ctimes10%5E%7B-31%7D%5Ctimes%286.0%5Ctimes10%5E%7B-9%7D%29%5E2%7D)
![E=6.709\times10^{-21}\ J](https://tex.z-dn.net/?f=E%3D6.709%5Ctimes10%5E%7B-21%7D%5C%20J)
(c). We need to calculate the energy of photon
Using formula of energy
![E = \dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
![E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B6.63%5Ctimes10%5E%7B-34%7D%5Ctimes3%5Ctimes10%5E%7B8%7D%7D%7B6.0%5Ctimes10%5E%7B-15%7D%7D)
![E=3.315\times10^{-11}\ J](https://tex.z-dn.net/?f=E%3D3.315%5Ctimes10%5E%7B-11%7D%5C%20J)
(d). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
![\lambda=\dfrac{h}{\sqrt{2mE}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2mE%7D%7D)
![E=\dfrac{h^2}{2m\lambda^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bh%5E2%7D%7B2m%5Clambda%5E2%7D)
Put the value into the formula
![E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B%286.63%5Ctimes10%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes9.1%5Ctimes10%5E%7B-31%7D%5Ctimes%286.0%5Ctimes10%5E%7B-15%7D%29%5E2%7D)
![E=6.709\times10^{-9}\ J](https://tex.z-dn.net/?f=E%3D6.709%5Ctimes10%5E%7B-9%7D%5C%20J)
Hence, This is the required solution.