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anyanavicka [17]

3 years ago

10

Which of the following would demonstrate an attraction between magnetic poles?

2 answers:

Korolek [52]3 years ago

8 0

Answer: The correct answer is C: North pole to a south pole.

Explanation: Opposite poles attract each other while alike poles repel one another.

hope this helps (:

vredina [299]3 years ago

5 0

"C. north pole to a south pole" would demonstrate an attraction between magnetic poles, since in the field of magnetics, opposites attract and similar forces repel one another.

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eimsori [14]

Can be absorbed and transformed into heat. i know this because i used to work as a constructor for lights and i had to go over a training on this subject.

8 0

3 years ago

Read 2 more answers

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

3 years ago

eimsori [14]

B. Unbalanced force

3 0

3 years ago

A system has two possible energy states, E0 and E1 (with E1 > E0). If the difference between these energy states grows (E0 an

Ugo [173]

Answer:

The number of particles in state E0 over the number of particles in state E1 will reduce

Explanation:

E0 represents the ground level state when all the particles have same energy level.

E1 represents excited state in which only a few particle reaches

E0 and E1 get further apart means that the energy difference between the two level increases.

Thus, the number of particles in state E0 over the number of particles in state E1 will reduce.

5 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago