PLEASE HELP!!! I have so much work left to do. PLEASE!!!

Why did you need a permanent magnet? What do you think would happen if you increased the

inysia [295]

Permanent magnets are important for their industrial uses especially when it comes to power generation and electric motors. The induction process for turbines and generators needs permanent magnets to turn mechanical motion into energy.

Exposing magnetized magnets to opposing external fields produces a demagnetizing effect. This effect is most apparent in permanent magnet materials having a non-linear demagnetization curve, such as Alnico. ... Other factors such as SHOCK and VIBRATION have very little effect on today's permanent magnet materials.

7 0

3 years ago

Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it

lesya692 [45]

Answer:

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

$\theta = \omega t$

now the position vector at a given time is

$r = Rcos\theta \hat i + R sin\theta \hat j$

now the linear velocity is given as

$v = \frac{dr}{dt}$

$v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}$

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$

6 0

3 years ago

3)

sasho [114]

As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. ... In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them. The temperature of a boiling liquid remains constant, even when more heat is added.

6 0

3 years ago

If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find

N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$

Original width = 'w'.

So, $l=2\frac{1}{2}w=\frac{5}{2}w$

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = $l+7=\frac{5w}{2}+7$

New width (w') = $w+7$

New perimeter (P) = 266 ft

Perimeter is given as:

$P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft$

Therefore, original width = 38 ft.

Original length is, $l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft$

Hence, the original dimensions of the building is 95 ft × 38 ft.

7 0

4 years ago

A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m

charle [14.2K]

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) $x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

$K_{eq} + U_{eq} = K_A + U_A\\\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\v = \sqrt{\frac{k}{m}}A$

where $\omega = \sqrt{k/m}$ and $\omega = 2\pi f$

Therefore,

$v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s$

Part B

The conservation of energy will be used again.

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}$

where $k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89$

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

$E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J$

Part D

The general equation of motion in simple harmonic motion is

$x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)$

where $\phi$ is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

$x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

5 0

4 years ago