Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.
1 answer:
Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s,
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.
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Answer:
E = 1.711 MeV
Explanation:
From the law of the conservation of energy:
where,
the kinetic energy of positron and electron = 1.2 MeV
Rest energy of the electron and the positron = 0.511 MeV
E = Energy of Photon = ?
Therefore,
<u>E = 1.711 MeV</u>
The angle above the horizontal of the motion of mass A + B after the collision. as well as others are mathematically given as
- vf/vA= 0.94
- 2vf^2 / 5vA^2 = 0.35
<h3>What is Collison equilibrium?</h3>
According to the collision hypothesis of chemistry, chemical reactions happen when two molecules or atoms collide.
Generally, the equation for Collison equilibrium is mathematically given as
For horizontal motion
mvAx+m2vAx = 2mvfx
For vertical motion
-mvAy+m2vAy = 2mvfy
vfy = 1/2 vAy
(b)
vf/vA= 0.94
(c)
2vf^2 / 5vA^2 = 0.35
In conclusion, The angle above the horizontal of the motion of mass A + B after the collision. as well as others
vf/vA= 0.94
2vf^2 / 5vA^2 = 0.35
Read more about collision theory
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