Ask question

Ask question

All categories

2 years ago

8

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

1 answer:

attashe74 [19]2 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:

You might be interested in

मारवतन गनुहास (What Is MKS Syste<br>Convert 5 solar days into second.)

mr_godi [17]

Answer:

5 Days to Seconds = 432000

Explanation:

7 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Determine the value of the information in scientific notation. The moon is approximately 3. 63 × 108 meters from Earth. What is

Usimov [2.4K]

Answer:

d

Explanation:

7 0

2 years ago

Will give brainliest, Pleaseee help!!!

puteri [66]

Answer:

below

Explanation:

1.1115 im not sure tho

5 0

2 years ago

How many 1140 nm long molecules would you have to line up end to end to stretch a distance of 158 miles?

dezoksy [38]

Answer:

221754385964.9123

Explanation:

Convert miles to nanometer

1 mile = 1.6 km

1 km = 1×10³×10³×10³×10³ nm

1 mile = 1.6×10¹² nm

So,

158 miles = 158×1.6×10¹² = 252.8×10¹² nm

Length of each molecule = 1140 nm

Number of molecules = Total length / Length of each molecule

$\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123$

There are 221754385964.9123 number of molecules in a stretch of 158 miles

3 0

3 years ago