Answer:
Explanation:
Parameters given;
Magnetic field intensity, B = 1.4 T
Speed of proton, v =
Mass of proton, m =
Charge of proton, q =
To find the acceleration of the proton, we first need to find the force exerted by the magnetic field on the proton:
F = q * v * B
F = * * 1.4
F =
This is the force exerted by the magnetic field on the proton. The force exerted by the proton has the same magnitude but an opposite direction as the force exerted by the magnetic field. Hence, = -
The force exerted by the proton is the product of its mass and acceleration. Hence, we can find its acceleration:
= ma
a = F/m
The magnitude, |a|, will be:
|a| =| |
|a| =
The magnitude of the acceleration of the proton is |a|
Answer:8 *10 ^-15 N= 8 fN
Explanation: In order to solve this we have take into account the following expression:
V=E*d the potential is equal to electric field multiply the separation between the plates.
The we have
E=V/d= 500 V/0.01 m=50000 N/C
then the force is given by
F=e*E = 1.6 * 10 ^-19 C*50000 N/C=8*10^-15 N
initial velocity = 0 = v₍i₎
final velocity = ? = v₍f₎
t = 12 sec
Acceleration = 4m/s²
First we have to find the distance d, for this we use the
formula,
D = v₍i₎t + 1/2at²
D = 0(12) + ½ (4)(12)²
Distance = d = 288 m
Now to find the Vf use the formula,
V₍f₎² = v₍i₎² + 2ad
V₍f₎² = (0)2 + 2(4)(288)
V₍f₎² = 2304
V₍f₎ = 48 m/s
so the velocity at the end of 12 sec is 48 m/s
Answer:
585 nm
Explanation:
The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:
where
m is the order of the maximum
is the wavelength
D is the distance of the screen from the slits
d is the separation between the slits
The distance between two consecutive bright fringes therefore is given by:
In this problem we have:
(distance between two bright fringes)
D = 2.0 m (distance of the screen)
d = 3.0 x 10−3 m (separation between the slits)
Solving for , we find the wavelength:
Answer:
v = √ k e²/m r
Explanation:
From the classical point of view the force between the nucleus and the electron is electrostatic
F = q E
q = e
we apply Newton's second law
F = m a
where the centripetal accelerations
a = v² / R
we substitute
e E = m v² / r
v² = ( e/m E r)
the electrioc field is
E = k q/r²
v² = e/m k e/r
v = √ k e²/m r