All categories

3 years ago

Do you think there is water on exoplanets

Physics

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

Explanation:

Some exoplanets may depending on the climate and vicinity from the sun.

You might be interested in

Distance versus Displacement Worksheet

Umnica [9.8K]

when we find the distance we will add all the blocks so

distance = 6+6+4

distance = 14blocks

when we find the displacement we will add and minus too

As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = 4blocks to the west

6 0

3 years ago

Read 2 more answers

Which is a more objective measurement, sound intensity or loudness, and why? 1. sound intensity is exactly same as loudness. 2.

fgiga [73]

The correct answer is:
2. sound intensity is a more objective and physical attribute of a sound wave because loudness can vary from person to person

indeed, sound intensity is a measurable quantity, and so it is objective, while loudness is the subjective perception of the sound level, so it varies from person to person.

7 0

3 years ago

Read 2 more answers

Which inference about Arthur is supported by the text?

yarga [219]

There is no picture

8 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

What 2 properties are used to classify radiation

ahrayia [7]

Large amount of energy is release

Temperature and pressure have no effect on the rate qt which radiation is emitted

6 0

3 years ago