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3 years ago

How do you make a electromagnet repel medal

Physics

1 answer:

Naddika [18.5K]3 years ago

8 0

Answer:

to make a magnet repel a metal, one must first understand the properties of a magnet. A magnet has two poles, a north pole and a south pole. When magnets are placed near each other, opposite poles attract and like poles repel one another.

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What does the principal quantum number determine?

morpeh [17]

<span>principal quantum number (n) </span>represents the relative overall energy of each orbital

Hope this helps!

4 0

3 years ago

If you were to apply a force of 15N to a medicine ball with a mass of 10kg, what would its acceleration be

tresset_1 [31]

One of the equations for force is: Force = mass × acceleration.

The force and mass are given and so:

F=15N

m=10kg

By plugging in these values we obtain:

$F=ma\\\\15=10(a)\\$

$\frac{15}{10}=\frac{10(a)}{10}\\\\a=\frac{15}{10}=1.5m/s^2$

So the acceleration is a=1.5m/s^2

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3 years ago

A car is driven 125.0 km due west then 65.0 km due south. What is the magnitude of its displacement?

kodGreya [7K]

pythagoras' theorem on right angled triangle. sides 125, 65

sqrt (125^2 +65^2)

5 0

3 years ago

During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth

Marta_Voda [28]

(a) $2.68\cdot 10^{-6} W/m^2$

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

P is the power

A is the area of the surface considered

For the waves in the problem,

$P=182 kW = 1.82\cdot 10^5 W$ is the power

The area is a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

So, the intensity is

$I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2$

(b) $5.9\cdot 10^{-7} W$

In this case, the area of the reflection is

$A=0.22 m^2$

So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:

$P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W$

(c) $8.7\cdot 10^{-18} W/m^2$

We said that the power of the waves reflected by the aircraft is

$P=5.9\cdot 10^{-7} W$

If we assume that the reflected waves also propagate over a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

which has an area of

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

Then the intensity of the reflected waves at the radar site will be

$I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2$

(d) $8.1\cdot 10^{-8} V/m$

The intensity of a wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the maximum value of the electric field vector

Solving the equation for $E_0$ ,

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m$

(e) $1.9\cdot 10^{-16} T$

The maximum value of the magnetic field vector is given by

$B_0 = \frac{E_0}{c}$

Substituting the values,

$B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T$

And the rms value of the magnetic field is given by

$B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T$

7 0

3 years ago

Make the following prefix conversions.<br> 0.001s =ms

Free_Kalibri [48]

Answer:

To convert a millisecond measurement to a second measurement, divide the time by the conversion ratio. The time in seconds is equal to the milliseconds divided by 1,000.

Explanation:

hope it helps

4 0

2 years ago

How do you make a electromagnet repel medal​

How do you make a electromagnet repel medal