A system consists of two spheres, of mass m and 2m, connected by a rod of negligible mass, as shown above. The system is held at
1 answer:
Answer:
Rod will remain horizontal all the time after release.
Explanation:
This is because net torque on the rod about any point in space is zero.
Let assume that distance between the two masses m and 2m is L.
Also m is situated at origin and in positive XY direction.
Then , Center of mass is at L()
COM =
So let us calculate net torque about hinged point which COM.
- Torque because of hinge force is zero because it passes from that point itself.
- Torque on 2m mass is 2mg(L/3) in nenegative Z direction.
- Torque on m mass is mg(2L/3) in positive Z direction.
As both torque are equal and opposite then net torque =0.
Thus it got balanced.
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We must write down laws of conservation of momenta and energy.
For the law of conservation of momenta will we will use two axes. One will be x-axis that will correspond to the east, and the other one will be y-axis corresponding to the north. Jack will be marked as 1 and Jill will be marked as 2.
Law of conservation of energy:
This will give us Jill's velocity after the colision.
Law of conservation of momenta:
We will use the second equation to get the angle at which the Jill is traveling:
When we plug all the number we get:
Please note that this is the angle below the x-axis.
For the law of conservation of momenta will we will use two axes. One will be x-axis that will correspond to the east, and the other one will be y-axis corresponding to the north. Jack will be marked as 1 and Jill will be marked as 2.
Law of conservation of energy:
This will give us Jill's velocity after the colision.
Law of conservation of momenta:
We will use the second equation to get the angle at which the Jill is traveling:
When we plug all the number we get:
Please note that this is the angle below the x-axis.
Answer:
3.4 x 10⁴ m/s
Explanation:
Consider the circular motion of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of electron = 9.1 x 10⁻³¹ kg
v = radial speed
r = radius of circular path = 2 mm = 0.002 m
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
For the circular motion of electron
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Consider the linear motion of the electron :
v' = linear speed
x = horizontal distance traveled = 9 mm = 0.009 m
t = time taken = =
= 4.5 x 10⁻⁷ sec
using the equation
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
Speed is given as
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s