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tatuchka [14]
2 years ago
11

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. Where are you
Physics
1 answer:
Marta_Voda [28]2 years ago
8 0

You find yourself in a place that is unimaginably <u>hot and dense</u>. A r<u>apidly changing</u><u> gravitational field</u><u> </u>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<u> 10^-44 seconds</u>. The smallest time interval, or "zeptosecond," that has so far been measured is <u>10^-21 seconds</u>. A photon traveling at the speed of light would need one Planck time <u>to traverse a distance of one </u><u>Planck length</u>.

<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 <u>four universal </u><u>physical constants</u> 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 <u>based on characteristics of nature</u>, 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:

brainly.com/question/23791066

#SPJ4

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The work done by \vec F along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,

\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r

I assume the path itself is a line segment, which can be parameterized by

\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath

with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is

\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

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