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Marta_Voda [28]
3 years ago
9

A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?

Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

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As a solid element melts, the atoms become __________ and they have __________ attraction for one another. more separated, more
galina1969 [7]
The complete statement is 

As a solid element melts, the atoms become more separated and they have less attraction for one another. 

Let me explain to you how this happens. In solid phase. Its molecules are arranged in a very compact manner that is why it takes a definite shape and volume. When it is heated, the kinetic energy of the molecules increases. This is characterized by more frequent collisions. The rise in temperature causes the molecules to move rapidly by vibrating. When it reaches an amount of energy that causes the solid to change phase, this is called the latent energy. The molecules break their form and move farther away from each other until it resembles that of a liquid melting. At this point, the molecules would have lesser attraction because of the distance between them.
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3 years ago
On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se
zhuklara [117]
It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.
3 0
3 years ago
Read 2 more answers
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
3 years ago
A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of
Roman55 [17]

Answer:

T=9.42Nm

Explanation:

From the question we are told that:

Radius r= 0.5 m

Current I= 3.0 A

Normal vector n=\frac{(2i - j +2k)}{3}

Magnetic field B= (2i-6j) T

Generally the equation for Area is mathematically given by

 A=\pi r^2

 A=3.1415 *0.5^2

 A=0.7853 m^2

Generally the equation for Torque is mathematically given by

 T=A(i'*B)

Where

 i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}

 X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]

 X\ component\ of\ i'*B=12

Therefore

 T=0.7853*12

 T=9.42Nm

7 0
3 years ago
Please help with this problem. Thank you.​
4vir4ik [10]

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

4 0
3 years ago
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