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3 years ago

How do magnetic forces repel or attract?

1 answer:

3 0

“Magnets are surrounded by an invisible magnetic field that is made by the movement of electrons, the subatomic particles that circle the nucleus of an atom”

“Every magnet has both a north and a south pole. When you place the north pole of one magnet near the south pole of another magnet, they are attracted to one another. When you place like poles of two magnets near each other (north to north or south to south), they will repel each other.”

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____ is the force that moves people to behave, think, and feel the way they do, resulting in behavior that is energized, directe

pogonyaev

Answer:

Motivation

Explanation:

Motivation is the force that directs one's behavior. This force is required for repeated actions. There are two types of motivation:

Intrinsic

This type of motivation is comes from the individual

Extrinsic

This type of motivation is comes from an external influence.

Both conscious and unconscious factors influence motivation.

5 0

3 years ago

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .

8 0

3 years ago

A truck is traveling at 2.0 m/s. It slows to a stop at a constant rate over 3.00 s. How far does the car travel during those 3.0

earnstyle [38]

Answer:

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

Explanation:

The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.

Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.

Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:

a ⇒ b

c ⇒ x

So: $x=\frac{c*b}{a}$

In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?

$distance=\frac{3 s*2 m}{1 s}$

distance= 6 m

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

8 0

3 years ago

A plane is flying due east with the velocity of 90m/s. The wind is blowing out of the north at 4m/s. What is the magnitude of th

Butoxors [25]

as we know that the velocity vectors are at right angles
magnitude = ?
hypotenuse of a right triangle.
v^2 = 90^2 + 4^2
v^2 = 8116
Taking the square root of both sides here we get,
v = 90.1 m/s
hope it helps

4 0

3 years ago

A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does

Gelneren [198K]

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

$COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}$

Here $Q_C$ is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, $T_C$ is the cold temperature and $T_H$ is the hot temperature. Recall use absolutes temperatures( $273.15+^\circ C$ ). Replacing and solving for W:

$W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J$

8 0

3 years ago