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2 years ago

Select the correct location on the image. Identify the magnetic north pole of Earth’s magnet.

Physics

1 answer:

OverLord2011 [107]2 years ago

8 0

The magnetic north pole of the earth's magnet is in the geographic south pole.

There are two magnetic and geographic poles each, north and south
The two geographic poles are the locations where the earth's axis of rotation passes through which is imaginary
The magnetic north and south poles are not the same as the geographic north and south poles
In a compass, the needle points to the magnetic north pole
By convention, the magnetic north pole corresponds to the geographic south pole
The magnetic south pole corresponds to the geographic north pole
The magnetic field lines of a magnet start from the magnetic north pole and end at the magnetic south pole

The magnetic north pole of the earth's magnet is the geographic south pole.

Learn more about earth's magnetism here:

brainly.com/question/3928159

#SPJ10

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Kipish [7]

Answer:

Velocity

Explanation:

"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."

^^This explanation is from physicsclassroom.com

3 0

3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e

My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

$PV =NkT$

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

$N = 1$

$k = 1.38*10^{-23}J/K$

$T = 27\°C = 27\°C + 273 = 300K$

$P = 1atm = 101325Pa$

Rearrange the equation to find V we have,

$V = \frac{NkT}{P}$

$V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}$

$V = 4.0858*10^{-26}m^3$

We know that length of a cube is given by

$V = L^3$

Therefore the Length would be given as,

$L = V^{1/3}$

$L = (4.0858*10^{-26})^{1/3}$

$L = 3.445*10^{-9}m$

Therefore each length of the cube is 3.44nm

7 0

3 years ago

Who was the creator of sports

Elden [556K]

The first Olympic Games were in Greece

8 0

3 years ago

An example of constant velocity

pashok25 [27]

Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.

4 0

3 years ago