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3 years ago

Identify the magnetic North Pole of Earth's magnet

Physics

2 answers:

Katen [24]3 years ago

8 0

Answer:

As we know that magnetic field lines always originated out from the north pole of magnet and then terminate at south pole of the magnet

So here in order to find out the magnetic north pole we can say that magnetic field lines will originated out from that pole

In order to find out that part of earth from which magnetic field lines originated out we have to suspend a bar magnet freely in the air.

Now the south pole of bar magnet is always attracted towards the North pole of Earth magnetic field

So when We freely suspend the bar magnet then its south pole always shows the position of North pole of Earth magnetic field.

So we can say that the Geographic south position of Earth is the magnetic North pole of Earth

AlladinOne [14]3 years ago

6 0

If i am reading this question right the answer should be the south pole

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Three identical very dense masses of 7500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

sukhopar [10]

Answer:

0.00354 (N)

Explanation:

Convert to metric system:

$x_1 = -100 cm = 1 m$

$x_2 = 420 cm = 4.2 m$

Formula for gravitational force:

$F_g = G\frac{mM}{s^2}$

where s is the distance between 2 bodies masses m and M

Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:

$F_g = F_{g1} - F_{g2}$

$F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}$

$F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})$

$F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2})$

$F_g = 5*10^{-7}(7500 - 425.17)$

$F_g = 5*10^{-7} * 7074.83$

$F_g = 0.00354 (N)$

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3 years ago

The type of waves that transmit cell phone messages are most likely

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Cell phones use RF waves from nearby cell towers. RF waves fall in between FM radio waves and microwaves

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3 years ago

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If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final<br>temperature?

Debora [2.8K]

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

$Q=mc\Delta T$

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17 K.

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3 years ago

What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc

Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

$Q= mc\Delta \theta$

where;

c is the specific heat capacity
Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

$Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J$

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

$C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC$

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

$C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC$

Specific heat capacity of the metal for the third trial;

$C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC$

Specific heat capacity of the metal for the fourth trial;

$C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC$

Specific heat capacity of the metal for the fifth trial;

$C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC$

Average specific heat capacity

$C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC$

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

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