It depends on the direction in which the shell is launched. The time can be anything from 3.6 seconds to never.
The combined momentum is 4000 kg m/s south
Explanation:
The total combined momentum of the two cars is given by the vector addition of the momenta of the two cars.
For this problem, we choose north as positive direction and south as negative direction.
The momentum of the first car travelling north is given by:

where
is the mass of the car
is its velocity
Substituting,

The momentum of the second car travelling south is given by:

where
is the mass of the car
is its velocity (negative because the car travels south)
Substituting,

And therefore, the combined momentum is

where the negative sign means the direction of the total momentum is south.
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Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
(1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):

The kinetic energy of the particle is also:
(2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:

the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:

B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:

the radius of the trajectory of the electron is 10.1 cm
The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has
1/2 (mass) (speed squared)
of kinetic energy when it lands, and zero when it stops.
So
1/2 (mass) (original speed squared)
is the energy it loses to friction in order to come to rest.
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m