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jeyben [28]
3 years ago
11

The center of mass of a 0.3 kg (non-uniform) meter stick is located at its 46-cm mark. What is the magnitude of the torque (in N

m) due to gravity if it is supported at the 25-cm mark? (Use g = 9.79 m/s2)
Suppose that a meter stick is balanced at its center(50 cm mark). A 0.27 kg mass is then positioned at the 20 cm mark. At what cm mark must a 0.30 kg mass be placed to balance the 0.27 kg mass?
Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

15cm

Explanation:

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Alexxx [7]

Answer:

The concept of zero net current in toroid occurs when we try to determine magnetic field induction outside toroid.

to determine magnetic field outside toroid, we consider a circular loop outside toroid as shown in figure.

to get the net enclosed current inside the loop, let us consider the crosssection AA' in toroid as shown in figure

as we see the crosssection AA' part of figure, we see that currents in the winding are in opposition directions at near to the loop

and far away from loop. Hence the concept of net current zero occurs in toroid when we try to determine

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Have a Good day...

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When you add heat energy the molecules move ______ and _____
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They move faster and move farther apart
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A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?
olchik [2.2K]

Answer:

m=5.78\times 10^{-3}\ g

n_e=1.935\times 10^{20} is the no. of electrons

Explanation:

Given:

  • quantity of charge transferred, Q=31\ C

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron 1.602\times 10^{-19}\ C

so,

n_e=\frac{Q}{e}

n_e=\frac{31}{1.602\times 10^{-19}}

n_e=1.935\times 10^{20} is the no. of electrons

  • Now if each water molecules donates one electron:

Then we require n=1.935\times 10^{20} molecules.

<u>Now the no. of moles in this many molecules:</u>

n_m=\frac{n}{N_A}

where

N_A=6.022\times 10^{23} Avogadro No.

n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}

n_m=3.213\times 10^{-4}\ moles

  • We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

n_m=\frac{m}{M}

where:

m = mass in gram

3.213\times 10^{-4}=\frac{m}{18}

m=5.78\times 10^{-3}\ g

7 0
3 years ago
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