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3 years ago

11

Your eyes see different wavelengths of light as different ________.

1 answer:

zaharov [31]3 years ago

5 0

Answer:

Its color

Explanation: I got it right

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Given that an atom of a semiconductor has a diameter of 2.18 å , what is the maximum number of moles that fit in the channel of

andrezito [222]

<span>Convert angstroms to nm for atom diameter 2.18/10=.218 nm. Divide diameter by length width and height. 63.6/.218=292 74.2/.218=327 275/.218=1261 Multiply these to get volume of atoms 120,037,500 Convert atoms to moles using Avogadro number 120,037,500/6.02*10^23=2*10^-16 moles</span>

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4 years ago

3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided

Nitella [24]

Answer:

hope fully it help s

3 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

3 years ago

A stone weighing 0.7 kilograms rolls down the inclined plane from position B to position A. Position A is located at sea level.

ale4655 [162]

This is the explanation

5 0

2 years ago