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3 years ago

What is the pressure at 5000 km below the surface of the earth

Physics

1 answer:

attashe74 [19]3 years ago

7 0

Answer:

The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.

Explanation:

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A student drops an egg from the roof of their house onto a trampoline. The egg feels a change in momentum of 2.2 kg^ * m/s in 1.

geniusboy [140]

Answer:

F = m a = m v / t where v is the change in velocity in time t

F = p / t since m v is equal to p

F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N

Or you can use the impulse equation

8 0

2 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Explain the relationship between speed, velocity and acceleration

Montano1993 [528]

Speed is the rate at which something covers a distance; velocity is the same but it takes into account whether it goes forwards or backwards; and acceleration is the rate of an increase in speed.

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4 years ago

A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true

Fantom [35]

Answer:

(d) not enough info

Explanation:

because it doesn't specify where the strings are attached

if it was the two ends of the rod then T1 would be equal to T2

6 0

3 years ago

A ________ detects the speed that a given device can handle and communicates with it at that speed

melisa1 [442]

Switch because look A switch detects the speed that given device can handle and communicates with it at that speed

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3 years ago