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Nookie1986 [14]
3 years ago
5

ILL GIVE BRAINLYEST

Physics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

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Chất rắn có tính dị hướng là vật rắn
kari74 [83]

Answer:

Anisotropy, in physics, the quality of exhibiting properties with different values when measured along axes in different directions. Anisotropy is most easily observed in single crystals of solid elements or compounds, in which atoms, ions, or molecules are arranged in regular lattices.

Explanation:

HOPE IT HELPS

6 0
2 years ago
How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?
ICE Princess25 [194]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

W = P\Delta V

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

W = P(0\ m^3)\\

<u>W = 0 J</u>

5 0
3 years ago
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
During which phase of the moon do neap tides occur?
Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

F = G\frac{m1\cdot m2}{r^{2}} (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

               

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

           

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0
3 years ago
Read 2 more answers
In the problem below, what is the student showing?
Phantasy [73]

Answer:

S<em>tudent showing the volume of cube</em>

                      V = a³

                       if a = 3 cm

  Then volume is a³ = 3 ×3 × 3

                                 = 27 cm³

6 0
3 years ago
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