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igor_vitrenko [27]
3 years ago
10

After which action would the concentration of a solution remain constant?(1 point)

Physics
1 answer:
Alika [10]3 years ago
5 0

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

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The diagram shows parts of a wave. A series of waves with an arrow passing through their centers. The highest point of one wave
Pachacha [2.7K]

Answer:

im not sure but i think its c

Explanation:

3 0
3 years ago
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A ball has a mass of 0.1 kg and an initial velocity of 20 m/s. The ball is given an acceleration of 30 m/s2 for 5 s. What is the
djverab [1.8K]

Answer:

acceleration = v-u /t

30- 20/5

= 10/5 = 2m/sec²

Force = mass * acceleration

Force = 0.1 * 2

Force = 0.2 Newton

6 0
3 years ago
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Can lamp that works on a 2.5 v work on a 1.12 v ?​
12345 [234]

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

5 0
3 years ago
The igneous rocks which were deposited on the surface and then cooled are known as __________.
mr Goodwill [35]
The igneous rocks which were deposited on the surface and then cooled are known as extrusive. These rocks are a result of a magma reaching the surface of the Earth which cools the magma quickly. Examples are rhyolite, basalt, obsidian and andesite.
6 0
3 years ago
A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The
cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

now for pure rolling condition we know that

v = R\omega

so we will have

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})

mgh = \frac{7}{10}mv^2

now we will have

v^2 = \sqrt{\frac{10}{7}gh}

v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}

v = 4.1 m/s

7 0
3 years ago
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