before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
1 answer:
The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for , we find:
So, the momentum of block B after the collision is -50 kg m/s.
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Answer:
Explanation:
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This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ =
Hence for a given w , θ depends upon g or weight .
Answer: 2.068*m
Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.
workdone= qV
energy = hc/λ
q=magnitude of an electronic charge= 1.602*
h= planck constant = 6.626*
c= speed of light =2.998*
v= potential difference= 6*
λ= wavelength=unknown
by making λ subject of formulae we have that
λ=
λ = 6.626* * 2.998*
/ 1.602*
* 6*
λ =
by doing the necessary calculations, we have that
λ = 2.068*m