Answer: 363 Ω.
Explanation:
In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:
Z = √((R^2 )+〖(XL-XC)〗^2) (1)
In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.
We are told that it has been set to 5.6 times the resonance frequency.
At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:
fo = 1/2π√LC = 286 Hz
So, we find f to be as follows:
f = 1,600 Hz
Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:
Z = 363 Ω
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x = 
x = 
let's calculate
x = 
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
2 someone's glasses could fall of because of the inertia
roller coaster is in motion but your glasses are opposing and resisting that motion due to inertia....
1 there is a possibility that electrons move from wheels to the rail and vice versa which also means that it is producing some kind of electrical energy
3 question can't answer
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
The value is 
Explanation:
From the question we are told that
The magnitude of the horizontal force is 
The mass of the crate is 
The acceleration of the crate is 
Generally the net force acting on the crate is mathematically represented as
Here 
is force of kinetic friction (in N) acting on the crate
So
=>
