Which type of magnet is created from an electromagnet

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0

2 years ago

In this case, lithium would be classified as a(n)

Black_prince [1.1K]

Lithium is to be considered as an ION

Atom is defined as its natural state where it will have all its electrons present in its shell

Since in this case Lithium is giving its electron so after giving the electron it will become an ion

Compound is the combination of two or more different atoms

Molecules is the combination of same type of atoms

So here answer must be

<em>C) ion</em>

5 0

3 years ago

Read 2 more answers

Why do sound waves move faster through the ground than through the air? a. Particles of matter are packed more loosely in the gr

lbvjy [14]

B. is it i just got done this in class like two weeks ago hope it helps

5 0

3 years ago

Read 2 more answers

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$