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Sholpan [36]
4 years ago
11

Take an electric field sensor and move it in a straight line, crossing the equipotential lines. Describe the relationship betwee

n the distance between the equipotential lines and the strength of the electric field.
Physics
1 answer:
alex41 [277]4 years ago
4 0

Answer:

 E = - dV / dx

Explanation:

The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.

The electric field and the potential are related by

          E = - dV / dx

therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.

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An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s a
Cerrena [4.2K]

Assuming constant speeds, the P-wave covers a distance <em>d</em> in time <em>t</em> such that

9000 m/s = <em>d</em>/(60 <em>t</em>)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = <em>d</em>/(60(<em>t</em> + 1))

Now,

<em>d</em> = 540,000 <em>t</em>

<em>d</em> = 300,000(<em>t</em> + 1) = 300,000 <em>t</em> + 300,000

Solve for <em>t</em> in the first equation and substitute it into the second equation, then solve for <em>d</em> :

<em>t</em> = <em>d</em>/540,000

<em>d</em> = 300,000/540,000 <em>d</em> + 300,000

4/9 <em>d</em> = 300,000

<em>d</em> = 675,000

So the earthquake center is 675,000 m away from the seismic station.

6 0
3 years ago
If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, how far away is b if a i
m_a_m_a [10]

If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, the far of b if a is 200 mpc away is   100 Mpc

<h3>Further explanation </h3>

The parsec (pc) is a unit of length used to measure the large distances to astronomical objects outside the Solar System.

Although parsecs are used for the shorter distances within the Milky Way, multiples of parsecs are also required for the larger scales in the universe such as Kiloparsecs (kpc) is for the more distant objects within and around the Milky Way, megaparsecs (Mpc) for mid-distance galaxies, and gigaparsecs (Gpc) for many quasars and the most distant galaxies.

If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, how far away is b if a is 200 mpc away? Remember that v = h_0 * d

galaxy A:  2(galaxy B's V) = H_0 * 200

galaxy B's V = H_0 * 100

galaxy B:  100 Mpc

<h3>Learn more</h3>
  1. Learn more about  the recession velocities brainly.com/question/12578064
  2. Learn more about galaxy brainly.com/question/3358187
  3. Learn more about mpc brainly.com/question/3191068

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  the recession velocities

Keywords:  galaxy,  the recession velocities, velocity, mpc, measure, far away

7 0
3 years ago
Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass h
mestny [16]
<span>Answer: sin(incidence)/sin(refraction) = n_refraction/n_incidence sin(50) / sin(x) = 1.5 / 1 sin(50)/1.5 = sin(x) sin(x) = 0.511 x = 30.71o B] 50 degrees, same as the angle going in. You can show that by reversing the steps in A. sin(30.7)/sin(x) = 1/1.5 C] The glass is 5 cm thick. The reference angle = 30.7o Tan(30.7) = displacement / thickness Tan(30.7) = x / 5 5*sin(30.7) = x x = 2.97 cm which is the displacement.</span>
4 0
4 years ago
What is a blockchain? Select one: K a. None of these b. A consensus network that enables a new payment system and a completely d
Furkat [3]
C. A shared public ledger on which entire bitcoin network relies
7 0
3 years ago
How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?
Rus_ich [418]

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be <u>23,34 meters per second squared</u>

8 0
3 years ago
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