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3 years ago

O How are electricity and magnetism related?

1 answer:

6 0

Magnetism is the product of a moving charged particle. We can have electricity without magnetism but we can not have magnetism without electricity.An electro magnet is made so that we have a soft metal core and electricity around it. A bar magnet is a normal magnet in bar shape with permanent magnetism.

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Please help me with the question below the best answer with an explanation will get brainliest

lesya [120]

Answer:

Explanation:

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6 0

2 years ago

Tiny cell structure that carry out specific functions within the cell

RSB [31]

Answer:

Organelles

Explanation:

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3 years ago

Definition These are the components of the reservoirs of oxygen that are exchanged in our environment.

andrew11 [14]

Answer:

Oxygen cycle

Explanation:

The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle

It suggests the movement of oxygen between the living and non-living parts.

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4 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

3 years ago

The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$