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GaryK [48]
3 years ago
12

Is the temperature and density of Earth's crust high or low? Why?

Physics
1 answer:
Setler [38]3 years ago
7 0

Answer:

The crust is the first layer of the earth. It is split up into two parts the continental crust, and the oceanic crust. The oceanic crust takes up 71% of the earths crust, and the other 29% of the crust is continental. The continental is made up of igneous rocks, and the oceanic crust is made up of sedimentary and basalt rocks. The continental crust is older than the oceanic crust, some of the rocks are 3.9 billion years old. The density average of the oceanic crust is 3g/cm. The average density of the continental earth is 2.7g/cm. The temperature of the crust is around 200-400 degrees Celsius. The crust is about 60 km thick under a continent and 5 km thick under the ocean. The crust is constantly moving.  The crust doesn't even make up 1% of the earth!  The crust is the layer were tectonic plates can be found.

Explanation:

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Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas
Westkost [7]

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

8 0
3 years ago
**100 POINTS**
lawyer [7]

Answer:

B

Explanation:

6 0
3 years ago
What is the final speed of a 60 kg boulder dropped from a 111 meter cliff
saveliy_v [14]

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

       Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

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3 years ago
The maximum oxygen uptake is known as the
grandymaker [24]
<span>The maximum oxygen uptake is known as the VO max.</span>
3 0
3 years ago
A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the
Oksanka [162]

The  average power produced by the soccer player is  710 Watts.

Given the data in the question;

  • Mass of the soccer player; m = 67kg
  • Energy used by the soccer player; E = 5100KJ = 5100000J
  • Time; t = 2.0h = 7200s

Power; P =\ ?

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

Power = \frac{Energy\ converted }{time}

We substitute our given values into the equation

Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W

Therefore, the  average power produced by the soccer player is  710 Watts.

Learn more: brainly.com/question/20953664

8 0
2 years ago
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