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2 years ago

We know that religion was important to healthcare practices in ancient Egypt the best support this is

Physics

1 answer:

Nonamiya [84]2 years ago

7 0

Assuming there are no choices, the answer might be the practice of calling upon gods and goddesses to help the sick.

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ფიზიკაში მჭირდება დახმარება ნაგლმა მასწავლებელმა მთელი თვე გააცდინა და როცა დაბრუნდა ეგრევე საკონტროლო ისეთი საკითხებით რომელიც

MakcuM [25]

Answer:

The photo is blocked cant see anything

Explanation:

3 0

2 years ago

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

3 years ago

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

4 0

2 years ago

How would the moon appear from Earth if the moon did not rotate?

andrey2020 [161]

It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon

6 0

2 years ago

Read 2 more answers

Suppose the entire population of the world gathers in ONE spot and everyone jumps at the sound of a prearranged signal. While ev

puteri [66]

Answer:

(b) Yes, the earth gains momentum but the change in momentum of the earth is much lesser compared to that of everyone in the air. The resistance to motion (inertia of the earth), which is a function of its mass is so great that the earth's acceleration is small in the given time frame.

Explanation:

From Newton's second law which can be stated mathematically as

F = m(v-u)/t = ma.

By Newton's law of gravitation, there is a force between the earth and everyone in the air. This force is responsible for the change in momentum of everyone in the air and this force gives them an acceleration equal to g = 9.80m/s². By Newton's law of gravitation and Newton's third law of motion, this force is also equal to the force exerted by everyone on the earth.

For this to be true,

F = M (everyone) ×a (everyone) = M(earth) × a (earth).

And

a (earth) = {M (everyone) ×a (everyone) }/M (earth)

Then

a (earth) must be lesser than a (everyone) since M(earth) >> M(everyone).

a = change in momentum/ time

Therefore the earth will have a much lesser change in momentum which is the reason we won't notice the earth's movement.

Thank you for reading.

6 0

2 years ago