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3 years ago

High pressure center of dry air

Physics

2 answers:

Virty [35]3 years ago

5 0

Anticyclone is the high pressure center of dry air

Nimfa-mama [501]3 years ago

3 0

A high pressure center of dry air is called an anticyclone

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The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

3 years ago

Which best describes the energy of a sound wave as it travels through a medium?

qaws [65]

Answer:

it increases

Explanation:

7 0

3 years ago

Read 2 more answers

Visible light falls into wavelength ranges of 400-700 nm, for which 1 m = 1 × 10 9 nm . The energy and wavelength of light are r

Paul [167]

Answer:

E = 3.54 x 10⁻¹⁹ J

Explanation:

The energy of the photon can be given in terms of its wavelength by the use of the following formula:

$E = \frac{hc}{\lambda}$

where,

E = energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light = 2.998 x 10⁸ m/s

λ = wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m

Therefore,

$E = \frac{(6.626\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{5.606\ x\ 10^{-7}\ m}\\$

E = 3.54 x 10⁻¹⁹ J

7 0

3 years ago

A typical adult human has a mass of about 70 kg.

Misha Larkins [42]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. FM GmMmr2
= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
= 2.40 x 10-3 N

b. FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%

3 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago