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tatuchka [14]
2 years ago
13

Convection is the transfer of energy by the motion of heated particles in a fluid. according to this information,which statement

best describes an example of convection
A a shirt heated by an iron

B a puddle heated by the sunlight

C a dark colored car heated by sunlight

D warm air rising
Physics
1 answer:
Anna [14]2 years ago
4 0

Answer:D warm air rising

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What should #1 be labeled?
aliya0001 [1]
B. North Pole because it’s pointing at the north pole, and not the others
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3 years ago
A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N
Pepsi [2]

Answer:

The mass of the planet  is 1.9407\times10^{27}\ kg

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon m=8.932\times10^{22}\ kg

We need to calculate the mass of the planet

Using Kepler’s third law

T^2\propto a^3

T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3

(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}

(8.932\times10^{22}+m)=1.94087\times10^{27}

m=1.94087\times10^{27}-8.932\times10^{22}

m=1.9407\times10^{27}\ kg

Hence, The mass of the planet  is 1.9407\times10^{27}\ kg

8 0
4 years ago
The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8
qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of  simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

If there is a distance between the lenses, the focal length will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}

\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}

\frac{1}{f_{2}} =\frac{-2}{24}

f_{2}= -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

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Method:

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