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2 years ago

What is seafloor spreading

Physics

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

Explanation:

The process by which molten materials add new oceanic crust to the ocean floor.

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Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Col

Vsevolod [243]

Answer:

Explanation:

From the question,

The kinetic energy of protons

K = 7eV.

Mass of proton and the speed of light are the given parameters.

Please kindly go through the files I attached for a step by step explanation of the answer required.

5 0

2 years ago

GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, a

SIZIF [17.4K]

The period of the wave is determined as 0.083 seconds.

<h3>What is period of a wave?</h3>

The period of a wave is the time taken by a particle of the medium to complete one vibration.

<h3>Period of the wave</h3>

The period of the wave is calculated as follows;

T = 1/f

where;

T is the period of the wave
f is frequency of the wave

T = 1/12

T = 0.083 seconds

Thus, the period of the wave is determined as 0.083 seconds.

Learn more about period of a wave here: brainly.com/question/18818486

#SPJ4

8 0

1 year ago

Which one do I press guys?

taurus [48]

Answer:

The Nucleus...

Explanation:

8 0

2 years ago

You are on a ParKour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal

Svet_ta [14]

The average speed would be 0.65 m/s, therefore the correct option is (A).

The average speed is calculated by the formula

Average speed= (total distance/ total time)

The total distance of the trip=9.5+3.5+15=28 m

The total time of trip=43 sec

Therefore the average speed=28/43=0.65 m/s.

8 0

3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago