Answer:
w = vR/3
Explanation:
The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.
From the principle of conservation of linear momentum
, we have
m*v = 2*m* Vcm
Where v = velocity of bullet, Vcm = velocity of wood
Hence, we have
Vcm = v2
Also, from the conservation of angular momentum about the centre of mass.
M*V*(R/2) = Ic*w - equation (I)
where Ic = moment of inertia and w = angular velocity
Ic for a ring is given by 
Ic of a bullet is given by 
Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives
Ic(system) = 
Substituting back into equation (I), we have
Hence, we obtain w =vR/3
w=v3R
When you catch the ball you act surprised, action- reaction
If I had to go with any of those answers, It would be A maybe D, But im not too sure on how to decide between them. Because Einstein mentioned the sun in his theory which has a very large mass <span> 1.989 x 10 with a exponent of 30 to be exact. Hope this helped though.</span>
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.
Ok, Rameses:
Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton
Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg
a). The force between two charges is
F = (9 x 10⁹) Q₁ Q₂ / R²
= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²
= ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²
= 8.05 x 10⁻⁸ Newton .
b). Centripetal acceleration =
v² / r .
A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)
= 7.7 x 10²² m/s² .
That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.
It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.
I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .
I'm going to leave that step for you. Good luck !
