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3 years ago

Which value is equivalent to 178 centimeters?

1 answer:

7 0

178 centimeters is exactly, precisely, the same length as ...

-- 0.00178 kilometer
-- 1.78 meters
-- 1,780 millimeters
-- 1,780,000 micrometers
-- 1,780,000,000 nanometers

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A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o

Liono4ka [1.6K]

Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, $d$ , and its volume (the volume of a sphere of radius r):
$M=dV=d ( \frac{4}{3} \pi r^3)= \frac{4}{3} \pi d r^3$

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
$g' = \frac{GM'}{r'^2}$ (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius:
$r'=2r$ (2)
and that its density is 2/3 of Earth's density:
$d'= \frac{2}{3} d$
so the mass M' of the new planet is, with respect to the Earth's mass:
$M' = d'V' = \frac{4}{3} \pi d' (r')^3 = \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) = \frac{16}{3} M$ (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
$g'= \frac{G( \frac{16}{3}M) }{(2r)^2}= \frac{GM}{r^2} \frac{4}{3} = \frac{4}{3}g$
And since $g=9.81 m/s^2$ , we find
$g'= \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2$

8 0

3 years ago

A sound whispered at one focus of a whispering gallery can be heard at the other focus. Two people are standing in a whispering

Temka [501]

Answer:75

Explanation:

6 0

3 years ago

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24

Anastasy [175]

Answer:

97 minutes and 12 seconds

Explanation:

Let x be the total time

From the first statement

$\frac{10}{100}x=24\times 60+18\\\Rightarrow 0.1x=1458\\\Rightarrow x=\frac{1458}{0.1}\\\Rightarrow x=14580\ seconds$

Total time of Dara's exercise is 14580 seconds

40 %

$\frac{40}{100}\times 14580=5832\ seconds$

$\frac{5830}{60}=97.2\ minutes=97+0.2\times 60\\ =97\ minutes\ and\ 12\ seconds=1\ hour\ 37\ minutes\ and\ 12\ seconds$

Time when she had completed 40% was 1 hour 37 minutes and 12 seconds

5 0

3 years ago

Earth’s orbital speed is approximately _____.

Aleks04 [339]

The answer that best fits the blank provided above is 30 KILOMETERS PER SECOND. The orbital speed of the Earth as recorded is 30 kilometers per second or <span>107000 km/h or</span> 18½ miles per second. The orbital period is 365 days which makes one year. Hope this helps.

5 0

3 years ago

Read 2 more answers

A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the?

soldier1979 [14.2K]

A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the Haldane effect

The less saturated hemoglobin is and the lower the partial pressure of oxygen in the blood is, the more readily hemoglobin binds to carbon dioxide. This is an example of the Haldane effect.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

To learn more about Haldane effect here

brainly.com/question/12420452

#SPJ4

4 0

1 year ago