As we know that acceleration is directly proportional to force, therefore as the force is doubled, acceleration gets doubled too.
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 net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.
A.
Waves can transfer energy but they can’t transfer matter