To determine an epicentral distance scientists consider the arrival times of what wave types

You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and

Marta_Voda [28]

You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and time. Gripped by these huge fluctuations, you notice that there is but a single, unified force governing the universe, you are in the early universe before the Planck time.

<h3>What is Planck time?</h3>

The Planck time is approximately 10^-44 seconds. The smallest time interval, or "zeptosecond," that has so far been measured is 10^-21 seconds. A photon traveling at the speed of light would need one Planck time to traverse a distance of one Planck length.

<h3>What is Planck length?</h3>

Planck units are a set of measuring units used only in particle physics and physical cosmology. They are defined in terms of four universal physical constants in such a way that when expressed in terms of these units, these physical constants have the numerical value 1. These units are a system of natural units because its definition is based on characteristics of nature, more especially the characteristics of free space, rather than a selection of prototype object, as was the case with Max Planck's original 1899 proposal. They are pertinent to the study of unifying theories like quantum gravity.

To learn more about Plank time:

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8 0

2 years ago

How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

Natasha_Volkova [10]

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

7 0

3 years ago

What is the Kinetic Energy of a 120 kg object that is moving with a velocity of 15 m/s

gizmo_the_mogwai [7]

Answer:

Kinetic Energy:120 x 15=1800

Explanation:

8 0

3 years ago

Read 2 more answers

Plz answer will give Brainliest answer to whoever does.

Jobisdone [24]

Answer:

A foot kicking ball.

An iPhone being charged

Explanation:

A foot kicking a ball is example of transportation of Kinetic energy

$\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2$

An iPhone being charged means the electronic energy is converted into machinery energy

6 0

3 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

so

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

3 years ago