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3 years ago

WILL MARK BRAINLIEST!!! Is string a ferromagnetic material?

Physics

1 answer:

Contact [7]3 years ago

4 0

Answer:

See, the string is made of nickel and steel (iron+carbon), materials that are ferromagnetic. That is, a magnet attracts guitar strings. When this ferromagnetic metal vibrates in the magnetic field of the pickup, that disturbs the red field lines which also cross through the coil (not shown).

Explanation:

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What can electromagnetic waves travel through?

Kisachek [45]

Electromagnetic waves do not require a medium to travel through. They can travel through empty space or matter.

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2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

Think about lifestyle choices that you make on a daily basis. Choose at least four lifestyle choices which you could change to i

eimsori [14]

A.The four lifestyle choices which I could change are:
1. The amount of physical exercise I engage in on a daily basis.
2, The number of hours I sleep.
3. The quantity of junk foods I consume daily and
4. Cultivation of the habit of spending time alone daily.
B. My days are always very busy, so I do not engage at all in physical exercise. Adding at least a thirty minute of physical exercise to my daily tasks will improve my health considerable. Due to the nature of my day, I usually eat on the move, thus I eat junk foods everyday. Eating healthy and balanced diet will go a long way in improving my health. I sleep for three hours only everyday, from 2 am to 5 am.This is because I have to leave my house very early in the morning and I don't usually get to sleep until 2 am in the morning. Spending more hours sleeping will improve my overall health. I want to start spending at least 30 minutes alone on a daily basis in order to meditate and to reflect over the events of the day. This will be a great boost to my mental health.

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2 years ago

At what common temperature will a block of wood and a block of metal both feel neither hot nor cold to the touch ?

Anastaziya [24]

When you touch an object and heat flows OUT of it, INTO your finger, you say the object feels hot.

When you touch an object and heat flows INTO it, OUT of your finger, you say the object feels cold.

If the object has the same temperature as your finger ... <em>around the mid-90s</em> ... then no heat flows in or out of your finger when you touch the object, and the object doesn't feel hot or cold.

6 0

3 years ago

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0

3 years ago