<span>For a particle moving in the +y and in a -x B field the force will be in the + z direction
Magnitude = q*v*B = 1.60 x 10^-19*1930*2.04 x 10^-3 = 0.63x10^-18N
a) the electric field creates a force = E*q in the +z direction
Fe = 4.27*1.60x10^-19 = 6.83 x 10^-19N
So F = 0.63 x 10^-18 + 6.83 x 10^-19 = 1.31 x 10^-18N
b) Now F = 0.63 x 10^-18 – 6.83 x 10^-19 = 0.5 x 10^-19N
c) Now add them as vectors so
F = sqrt((0.63 x 10^-18)^2 + (6.83 x 10^-19)^2) = 0.86 x 10^-18N</span>
In this item, it is assumed that what is being asked is the final velocity and direction after collision. This can be answered through the conservation of momentum.
Momentum is the product of the mass and the velocity,
M = m x v
If we take east as positive then, west would be negative.
Through the conservation,
(1050)(-15 m/s) + (6320 kg)(14 m/s) = (1050 + 6320) x v
The value of v from the equation is 9.87 m/s. Since it is positive, the direction is east.
Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2
Answer:
v = 2917.35 m/s
Explanation:
let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.
at any point in the orbit the forces acting on the satelite are balanced such that:
Fc = Fg
mv^2/r = GmM/r^2
v^2 = GM/r
v = \sqrt{GM/r}
= \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}
= 2917.35 m/s
Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.