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Lapatulllka [165]

3 years ago

11

PLEASE HELP!

1 answer:

lora16 [44]3 years ago

5 0

Answer:

3. Has no moons

2. One of the rocky planets

1. The moon has no atmosphere.

Explanation:

Mars moons [2]: Deimos, Phobos; Jupiter is considered a <em>Jovian</em><em> </em><em>[</em><em>Gas</em><em> </em><em>Giant</em><em>]</em><em> </em><em>planet</em><em>;</em><em> </em>our moon has a very thin atmosphere.

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Please help have due very soon?thank you

Andru [333]

The answer is :78 I think

8 0

3 years ago

If a car travels 60 mph for a distance of 180 miles, how much time<br> did it take?

jolli1 [7]

Answer:

3 hours

Explanation:

180 divided by 60 (mph means miles per hours by the way)

6 0

3 years ago

Newton's laws of motion are true on earth and space.

frez [133]

Answer:

False

Explanation:

It is a common misunderstanding that objects in space have no weight. If that were true, they would just float away from the Earth, the Sun and the other planets. Objects in low Earth orbit experience about 90% of the weight that they feel on the surface of the Earth.

8 0

2 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind

lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = $\frac{velocity of light in air/vacuum}{velocity of light in substance}$

velocity = $\frac{distance}{time}$

The time for travel is kept constant for all mediums.

refractive index (n) = $\frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}$

distance in medium = $\frac{distance in vacuum}{refractive index of medium}$

$S_{medium} = \frac{S_{vacuum} }{n_{medium} }$

For water, n= 1.33

$S_{water} = \frac{1 }{1.33}$

$S_{water} = 0.752m$

For glass, n=1.5

$S_{glass}= \frac{1 }{1.5}$

$S_{glass} = 0.667m$

For diamond, n= 2.42

$S_{diamond} = \frac{1 }{2.42}$

$S_{diamond} = 0.413m$

3 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago