Answer:6.0×10^5m/s
Explanation:
According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.
After their collision, the two bodies will move with a common velocity (v)
Momentum = mass × velocity
Let m1 be the mass of the proton = m
Let m2 be the mass of the alpha particle = m2
Let v1 be the velocity of the proton = 3.0×10^6m/s
Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).
Using the law,
m1v1 + m2v2 = (m1 + m2)v
m(3.0×10^6) + 4m(0) = (m + 4m)v
m(3.0×10^6) = 5mv
Canceling 'm' at both sides,
3.0×10^6 = 5v
v = 3.0×10^6/5
The common velocity v = 6.0×10^5m/s
Will reduce 4 times.
So called LUG (Law of Universal gravitation).
Answer:
0.435atm
Explanation:
cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.185 m3 at a pressure of 0.740 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.315 m3. If the temperature remains constant, what is the final value of the pressure?
Given
Initial pressure P1= 0.740atm
Initial volume V1= 0.185 m3
Final pressure P2= ?
Final volume V2= 0.315 m3
At constant temperature, the pressure of a syste is inversely proportional to volume, by Boyles law then
P1V1=P2V2
P2=P1V1/V2
=(0.185*0.740)/0.315
0.1369/0.315
= 0.435atm
Therefore, final pressure is 0.435atm
V = final velocity
u = initial velocity
a = acceleration
s = displacement
Take downward direction as positive.
v^2 = u^2 + 2as
v^2 = 0.8^2 + 2(1.6)(5.5)
v^2 = 18.24
v = 4.27 m/s
