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EleoNora [17]
3 years ago
6

II. Magnetic fields Magnets and magnetic fields EM 115 We have observed that magnets interact even when they are not in direct c

ontact. In electrostatics we used the idea of an electric field to account for the interaction between charges that were separated from one another. For magnetic interactions, we similarly define a magnetic field. A. Obtain a compass from a tutorial instructor. I. Use the compass to explore the region around a bar magnet. Describe the behavior of the compass needle both near the poles of the magnet and in the region between the poles.
Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Solution :

We all know that a bar magnet have two poles, the north pole and the south pole. These poles interacts with each other. The ends of the magnets having similar poles will push each other away while the poles with like charges will pull each others towards it.

The compass needle is also a magnet having south polarity as well as north polarity. When the compass needle is close to the bar magnet, it is opposite to the poles or along the poles. The compass needle shows the direction or is pointed towards the north. So when the compass needle is placed near the north pole of the bar magnet, the pointer of the compass needle points towards the north, i.e. it gets deflected because of he like charges. And when it is placed near the south pole of the magnet, it gets attracted towards it and is pointed towards the pole.

Now as we move the compass needle from the poles to the region that is between the poles, the compass needle pointer points towards the north direction every time. It show a deflection always. If we place the magnetic lines, we will see that the magnetic lines will exit from the north poles and enters the south pole of the bar magnet.  

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antiseptic1488 [7]
<h3><u>Answer;</u></h3>

volume = 6.3 × 10^-2 L

<h3><u>Explanation</u>;</h3>

Volume = mass/density

Mass = 0.0565 Kg,

Density = 900 kg/m³

             = 0.0565 kg/ 900 kg /m³

             = 6.3 × 10^-5 M³

but; 1000 L = 1 m³

Hence, <u>volume = 6.3 × 10^-2 L</u>

8 0
3 years ago
Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y
Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

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8 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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For point c:

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5 0
3 years ago
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mestny [16]

Answer:

5.02 m

Explanation:

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Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

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H = 196.5884×0.5/19.6

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8 0
3 years ago
A force which resist the relative motion of two bodies in direct contact is called _______
Elena-2011 [213]

Frictional Force is the answer..

Hope it helps

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