Answer:
S = V t - 1/2 a t^2 a here is negative due to deceleration
S = 4 * 8 - 1/2 * 64 = 0
The object just returns to its original starting point
It started out at 8 m/s which was reduced to 0 m/s after 4 sec
In 4 more seconds it was moving at -8 m/s, just the opposite of where it started
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
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 !
