When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t
V = f(wavelength)
22.0 = 0.0680 (wavelength)
wavelength = 323.52 m
Answer:
The answer is explained below.
Explanation:
All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.
So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.
<em>Rp - R</em>
<em>Rq - 2R</em>
The linear acceleration is
α2 - Rα
So, the linear acceleration of Q is twice as great as the linear acceleration of P.
The speed of the particle when it is in the circular motion depends on the radius of the particle.
In this case, the speed of point Q is twice the speed of point P.
