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mote1985 [20]
3 years ago
9

9. If you leave a paperclip stuck to a magnet long enough, the paperclip can become magnetic. Explain how this happens. Include

discussion of the “tiny magnet” property of atoms and domains.
Physics
1 answer:
Virty [35]3 years ago
4 0

Answer:

Each atom acts as a tiny magnet, when the magnetic field of each of the tiny magnets are aligned together, it gives a north pole and south pole. This is brought about due to prolonged exposure to a permanent magnet, which has caused the alignment of the atoms' individual magnetic fields to attain polarity. If the paperclip is made of iron, it can be magnetized and demagnetized easily. Once the paperclip is removed from the magnet, it will lose its magnetic properties.

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Why would knowing the characteristics of circuits be important in designing electrical circuits?
blondinia [14]
The characteristic of a circuit actually indicate how the circuit functions. When one designs a circuit they have a specific function in mind and must know how to combine components in order to fulfill this functions.
8 0
3 years ago
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Briefly explain how an electric motor works.
NeTakaya
You put electricity<span> into it at one end and an </span>axle<span> (metal rod) rotates at the other end giving you the power to drive a machine of some kind. 


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3 0
3 years ago
HELPPPP
Arisa [49]

Answer:

A. nuclear fusion reactions

C. it's still hot from the big bang

Explanation:

The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:

  • Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
  • Some heat accreted during the early formation of the earth and has not been lost till this day.
  • Heating due to friction

These are some of the sources of the earth's internal heat.

3 0
2 years ago
A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta
Tomtit [17]

Answer:

(a) 185 N/m

(b) 3.083 kg

Explanation:

(a)

Using,

E = 1/2ke²....................... Equation 1

Where E = work done to compress the spring, k = spring constant of the spring, e = compression of the spring.

make k the subject of the equation

k = 2E/e²............... Equation 2

Given: E = 3.7 J, e = 0.20 m

Substitute into equation 2

k = 2(3.7)/0.2²

k = 185 N/m.

(b)

Using,

F = ma.............. Equation 2

Where F = force applied to the spring, m = mass attached to the spring, a = acceleration of the spring.

But from hook's law,

F = ke................. Equation 3

substitute equation 3 into equation 2

ke = ma

make m the subject of the equation

m = ke/a................ Equation 4

Given: k = 185 N/m, e = 0.2 m, a = 12 m/s²

Substitute into equation 4

m = 185(0.2)/12

m = 3.083 kg

3 0
3 years ago
Read 2 more answers
A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a
scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
T_h = \frac{T}{2}

Now, we can determine the velocity at which the can was launched at using the following equation:
v_f = v_i + at

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}

***vsinθ is the vertical component of the velocity.

Solve for 'v':
vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}

Now, recall that:
W = \Delta KE = \frac{1}{2}m(\Delta v)^2

Plug in the expression for velocity:
W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}

B.

We can use the same process as above, where T' = 2T and Th = T.

v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}

C.

The work done in part B is 4 times greater than the work done in part A.

\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}

4 0
2 years ago
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