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3 years ago

What makes human travel to the other planets in our solar system difficult?

Physics

1 answer:

inysia [295]3 years ago

4 0

Answer:

Taking water and oxygen from Earth is rather expensive and hard to do.

Explanation:

Without water and oxygen, travel to other planets is basically impossible

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While eating a peanut butter and jelly sandwich peanut butter drips on Dana shirt she tries to remove the peanut butter from her

MakcuM [25]

Answer:

The water does not remove the peanut butter because the peanut butter is thicker substance than the water is, therefore it would not remove immediately.. It could also be because the brown dye or color in the peanut butter color or the peanut butter itself has already gone through her clothing Dana would have to remove it as much as she could with a towel water and some soap then throw it to the washer afterward.

Hope this helps, good luck:)

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3 years ago

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MissTica

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3 years ago

Describe the relationship between a compass and a magnet. Use C.E.R while answering. Your CER should have a one-sentence claim,

Allushta [10]

Answer:

Magnets interfere with compasses because compasses use the magnetic field to locate NESW (north east south west)

Explanation:

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3 years ago

Read 2 more answers

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

8 0

3 years ago

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

3 years ago