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3 years ago

Which actions most likely cause the domains in a ferromagnetic material to align?

Physics

2 answers:

Alexxx [7]3 years ago

8 0

Answer:

A ferromagnetic material is a temporary magnet. The domains in a ferromagnetic material are randomly arranged. Under certain actions, the domains align in a particular direction and the material acts as a magnet. The actions that can cause alignment of domains in a ferromagnetic material are:

rubbing the material against a magnet would cause the alignment of domains in the same direction as of the magnet.
passing electricity around the material would generate magnetic field which would cause domains to align along the direction of the field.
placing the material near a strong magnet would cause the alignment of domains in the direction of the field generated by the strong magnet.

Other actions like heating the material, placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.

patriot [66]3 years ago

8 0

Answer:

-rubbing the material against a magnet

-passing electricity around the material would generate magnetic field

-placing the material near a strong magnet

Explanation:

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3 years ago

A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

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4 years ago

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Suppose you want to test the hypothesis that plants grow taller when they receive more sunlight. Identify and independent variab

ollegr [7]

The result that should be established is in the form
y = f(x)
where x, the amount of sunlight is the controlled (independent) variable,
y = height (growth) that corresponds to the amount of sunlight. Therefore y depends on x.

Clearly,
x, the amount of sunlight is the independent variable. It can be controlled.
y, the measured amount of growth is the dependent variable.

Answer:
The independent variable is the amount of sunlight.
The dependent variable is the growth.

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4 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

Give a specific example of a scalar or vector quantity. Explain why thus type of quantity is used to describe this event.

dlinn [17]

Example of scalar: speed. Example of vector: velocity

Explanation:

In physics, there are two types of quantities:

- Scalar: a scalar quantity is a quantity having only magnitude, so it is just a number followed by a unit. Examples of scalar quantities in physics are:

Mass

Time

Speed

- Vector: a vector quantity is a quantity having both a magnitude and a direction. Examples of vector quantities in physics are:

Force

Acceleration

Velocity

The two types of quantities can be used in the same event, but in a different way. One of the most common example is the difference between speed and velocity.

In fact, let's consider an object moving in a uniform circular motion: it means that it is moving in a circle at a constant speed. The speed of the object measures only how fast the object is moving, but without telling anything about its direction of motion. The velocity, viceversa, also takes into account the direction of motion, and exactly for this reason, the velocity in a uniform circular motion is not constant, because the direction (it is a vector) is constantly changing. So, in a uniform circular motion, the speed is constant but the velocity is not.

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