Answer:
Explanation:
Let v be the velocity acquired by electron in electric field
V q = 1/2 m v²
V is potential difference applied on charge q , m is mass of charge , v is velocity acquired
2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²
v² = 844 x 10¹²
v = 29.05 x 10⁶ m /s
Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .
Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron
= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶
= 79.02 x 10⁻¹³ N .
Minimum force will be zero when electron moves along the direction of magnetic field .
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
