A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block and the embedded bullet land at a 4.0m from the base of the port. Find the initial velocity of the bullet.

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

V=

M+m

mv

Explanation:

please mark me as the brainliest answer and please follow me for more answers to your questions..

real, smaller and upside-down.   so the answer should be b.

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m