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2 years ago

How are the magnetic domains of a magnet different from the domains of an ordinary piece of metal?

1 answer:

6 0

Answer:

In a magnet, the domains all point in the same direction; in an ordinary piece of metal, they're all jumbled up.

Explanation:

In a magnet, the domains all point toward the north pole; in an ordinary piece of metal, they all point to the south pole.

Side note:
Hope this helps!
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

What happens to a wave when it moves from one medium to another?

Nina [5.8K]

One side of the wave changes speed before the other side, causing the wave to move

3 0

3 years ago

A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?

harina [27]

Answer:

impulse = 8820 kg· $\frac{m}{s}$ or 8820 N·s

Explanation:

Impulse J is equal to the average force $F_{av}$ multiplied by the elapsed time Δt or in equation form, J = $F_{av}$ Δt

As long as your force of 450 N is constant then that value is your average force $F_{av}$ and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg· $\frac{m}{s}$ or 8820 N·s.

6 0

2 years ago

A piano emits frequencies the range from a low of about 28 Hz to a high of about 4200 Hz. Find the range of wavelengths in air a

Step2247 [10]

V=wave velocity , f= frequency, λ=wavelength 
Use it to find corresponding wavelengths for f=28 Hz 
λ= v/f= 337/28=12.036 m

for f=4200 Hz 
λ= v/f=337/4200= 0.08 m 
So max. wavelength is 12.036 m and 
Min Wavelength is 0.08 m 
So the range is between .08 m and 12.036 m
Hope this helps.

4 0

3 years ago

You forgot to put what IRITONCF is unscrambled

Lesechka [4]

I'm Pretty sure it's FRICTION

4 0

3 years ago