The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W
1 answer:
You might be interested in
v = final velocity
u = initial velocity
t = time taken
the acceleration of the cyclist is
approximately 3.33 m/s^2
the acceleration of the car is
5.0 m/s^2
The time interval for which the proton remains in the field is -
Δt = .
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine time interval during which the proton is in the field.
<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s
Magnetic field intensity (B) = 0.3 T
Leaving velocity (u) = - 20 j m/s
Now -
The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -
d = =
Since, the radius of circular path is not given, we will assume it R.
Therefore, time for which proton remained in the field is -
t =
Hence, the time interval for which the proton remains in the field is -
Δt =
To solve more questions on Force on charged particle, visit the link below-
#SPJ4
The time the truck must apply the given force to increase its speed to given value is 5 s.
The given parameters;
- <em>applied force, F = 600 N</em>
- <em>mass of the truck, m = 1,500 kg</em>
- <em>speed of the truck, v = 2 m/s</em>
The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>
F = ma
Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.
Learn more here:brainly.com/question/1988795