A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight
, level track where it couples with a flatcar of smaller mass, originally at rest, so that the two cars then roll together without friction. Consider the two cars as a system from the moment of release of the boxcar until both are rolling together. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Next, consider only the process of the boxcar gaining speed as it rolls down the hump. Consider the boxcar and the Earth as a system. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Finally, consider the two cars as a system as the boxcar is slowing down in the coupling process. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved?
1 answer:
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F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Answer:
speed of electrons = 3.25 × m/s
acceleration in term g is 3.9 × g.
radius of circular orbit is 2.76 × m
Explanation:
given data
voltage = 3 kV
magnetic field = 0.66 T
solution
law of conservation of energy
PE = KE
qV = 0.5 × m × v²
v =
v =
v = 3.25 × m/s
and
magnetic force on particle movie in magnetic field
F = Bqv
ma = Bqv
a =
a =
a = 3.82 × m/s²
and acceleration in term g
a =
a = 3.9 × g
acceleration in term g is 3.9 × g.
and
electron moving in circular orbit has centripetal force
F =
Bqv =
r =
r =
r = 2.76 × m
radius of circular orbit is 2.76 × m