Answer:
When the magnetic field is tilted so it is no longer perpendicular to the page.
When the magnetic field gets stronger.
When the size of the loop decreases.
Explanation:
According to the Faraday-Lenz law, the change of the magnetic flux over time causes an induced current, this flux is given by:
![\Phi_B=BScos\theta](https://tex.z-dn.net/?f=%5CPhi_B%3DBScos%5Ctheta)
Therefore, there will be a variable magnetic flux, when the magnitude of the magnetic field (B) changes over time, when the area of the loop (S) changes over time and / or when the angle (
) between the field and the surface vector changes over time.
According to the general rules and basic knowledge of physics, without any doubds I can say that a mole of red photons of wavelength 725 nm has [D] 165 kj of energy. I converted <span> a wavelength into energy in that way :
</span>
![(E)= 6.023 x 10 ^23 x 6.625 x 10 ^ -34 x3 x 10^8/725x10^-9](https://tex.z-dn.net/?f=%28E%29%3D%206.023%20x%2010%20%5E23%20x%206.625%20x%2010%20%5E%20-34%20x3%20x%2010%5E8%2F725x10%5E-9)
=
![=0.1651 x10 ^ 6 J = 165 Kj.](https://tex.z-dn.net/?f=%3D0.1651%20x10%20%5E%206%20J%20%3D%20165%20Kj.)
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</span>
Answer:
That is true. They share atoms with each other. I hope this helps. Comment if you have any question
Explanation:
Answer:
The cylinder’s total kinetic energy is 1.918 J.
Explanation:
Given that,
Mass = 4.1 kg
Radius = 0.057 m
Speed = 0.79 m/s
We need to calculate the linear kinetic energy
Using formula of linear kinetic energy
![K.E_{l}=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=K.E_%7Bl%7D%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2](https://tex.z-dn.net/?f=K.E_%7Bl%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes4.1%5Ctimes%280.79%29%5E2)
![K.E_{l}=1.279\ J](https://tex.z-dn.net/?f=K.E_%7Bl%7D%3D1.279%5C%20J)
We need to calculate the rotational kinetic energy
![K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2](https://tex.z-dn.net/?f=K.E_%7Br%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20mr%5E2%5Ctimes%28%5Cdfrac%7Bv%7D%7Br%7D%29%5E2)
![K.E_{r}=\dfrac{1}{4}\times m\times v^2](https://tex.z-dn.net/?f=K.E_%7Br%7D%3D%5Cdfrac%7B1%7D%7B4%7D%5Ctimes%20m%5Ctimes%20v%5E2)
![K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2](https://tex.z-dn.net/?f=K.E_%7Br%7D%3D%5Cdfrac%7B1%7D%7B4%7D%5Ctimes4.1%5Ctimes%280.79%29%5E2)
![K.E_{r}=0.639\ J](https://tex.z-dn.net/?f=K.E_%7Br%7D%3D0.639%5C%20J)
The total kinetic energy is given by
![K.E=K.E_{l}+K.E_{r}](https://tex.z-dn.net/?f=K.E%3DK.E_%7Bl%7D%2BK.E_%7Br%7D)
![K.E=1.279+0.639](https://tex.z-dn.net/?f=K.E%3D1.279%2B0.639)
![K.E=1.918\ J](https://tex.z-dn.net/?f=K.E%3D1.918%5C%20J)
Hence, The cylinder’s total kinetic energy is 1.918 J.