Answer:
It depends on how you use the spoon...
if you are keeping one end of the spoon and pressing other end, the force you provide is supported by the force due to gravity... Hence it is easy to open this way :)
F + G ... Where F is the force you provide and G is the force due to gravity.
Answer:
Explanation:
Because when the electric bulb at night switched It emitted the light this light rays forms an image on our retina hence the bulb become visible at night
2 ans
When an electric bulb is switched on, the electricity is allowed to pass in its circuit and is converted into light energy. Thus, a bulb glows when switched on and the eyes perceives the light energy in the visible spectra which is given out by the bulb. Hence, the bulb becomes visible
The different types of radiation in electromagnetic spectrum are compared by the amount of energy found in the photons.
Radio waves have photons with low energies, microwave photons have a little more energy than radio-waves. Infrared photons still have more energy, then comes visible, ultraviolet, x-rays and the most energetic of all, gamma rays.
The energy associated with electromagnetic radiation is proportional to frequency and inversely proportional to wavelength. So, electromagnetic waves with shorter wavelengths have more energy.
On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light. On the other end of the spectrum are gamma rays with wavelengths billions of times smaller than those of visible light.
To know more about electromagnetic spectrum:
brainly.com/question/27839167
#SPJ4
Answer:
![v = 1.42 m/s](https://tex.z-dn.net/?f=v%20%3D%201.42%20m%2Fs)
Explanation:
While mass is falling downwards there is no frictional loss so here we can use mechanical energy conservation
So change in gravitational potential energy = gain in kinetic energy of the system
![mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2)
for uniform cylinder we will have
![I = \frac{1}{2}MR^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7DMR%5E2)
now we have
![mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh = \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%29%28%5Cfrac%7Bv%7D%7BR%7D%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh = \frac{1}{4}Mv^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B4%7DMv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh = (\frac{M}{4} + \frac{m}{2})v^2](https://tex.z-dn.net/?f=mgh%20%3D%20%28%5Cfrac%7BM%7D%7B4%7D%20%2B%20%5Cfrac%7Bm%7D%7B2%7D%29v%5E2)
now we have
![v^2 = \frac{mgh}{(\frac{M}{4} + \frac{m}{2})}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7Bmgh%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%20%2B%20%5Cfrac%7Bm%7D%7B2%7D%29%7D)
![v^2 = \frac{40(9.81)(0.50)}{(\frac{310}{4} + \frac{40}{2})}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B40%289.81%29%280.50%29%7D%7B%28%5Cfrac%7B310%7D%7B4%7D%20%2B%20%5Cfrac%7B40%7D%7B2%7D%29%7D)
![v = 1.42 m/s](https://tex.z-dn.net/?f=v%20%3D%201.42%20m%2Fs)