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2 years ago

How is the earth a cracked hard boiled egg

Physics

1 answer:

xz_007 [3.2K]2 years ago

6 0

Answer:

the earth isn't a cracked hard boiled egg, but I'd guess they mean layers. A thin shell/crust and the yolk and whites of the egg represent layers of the earth

Explanation:

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Part A: Explain why x = 5 makes 4x − 1 ≤ 19 true but not 4x − 1 < 19. (5 points) Part B: What value from the set {6, 7, 8, 9,

Elis [28]

Answer:

A: ≤ means less than OR equal to. < only means less than

B: 9

Explanation:

A: Because it would equal 19, and 19 is EQUAL than 19. 4(5) - 1 would equal 19, which is equal to 19, and not less than. ≤ means less than or equal to. < means less than. So its not true.

B: 47 - 2, 45. Then 5 x 9 equals 45. So 5 x 9 equals 45, then add 2 would equal 47.

Hope this helps <3

5 0

2 years ago

Calculate the magnitude of the gravitational force between Goku with a mass of 62 kg and King Kai’s planet with a mass of 1.458x

Brilliant_brown [7]

Answer:

6227.866 N

Explanation:

F = G . m(goku) . m(planet) / d²

F = 6.674 x 10-¹¹ x 62 x 1.458 . 10¹⁵ / 31²

F = 6227.866 N

7 0

2 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

2 years ago

A ship, carrying freshwater to a desert island in the caribbean, has a horizontal cross-sectional area of 2800 m2 at the waterli

Alex17521 [72]

Due that we already know the horizontal cross-sectional area of the ship, which is 2800 m2 and we are going to understand that value keeps constant for the whole 9.5 of height of the ship from the waterline till the new waterline after unloading, then we just need to calculate the volume as follows: V = A * H , where V is volume, A is area and H is height V= 2,800 * 9.5 = 26,600 m3 So this volum of 26,600 cubic meters is the volum of freshwater delivered in the island.

3 0

2 years ago

If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

Given the following the details;