Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Answer:
When an object moves in a straight line with a constant acceleration, you can calculate its acceleration if you know how much its velocity changes and how long this takes.
The formula is,
Acceleration = change in velocity / time taken
The equation for acceleration can also be represented as:
a = (v-u) \ t
The change in velocity v – u = 5 – 0 = 5 m/s.
The acceleration = change in velocity ÷ time = 5 m/s ÷ 2 s = 2.5 m/s^2
I don't think so as long as you make it apparent that the information comes the same source. So citing over and over again is unnecessary as long as it's clear that the information is from the same website or source. If you can't make it clear that they are from the same website source, it would a safe choice to continue to cite to avoid allegations of plagiarism.
