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3 years ago

To what extent do covalent compounds conduct electricity?

Physics

2 answers:

Snezhnost [94]3 years ago

7 0

Answer

I would say the answer is Never

Explanation:

Hope this helps - good Luck ^w

vagabundo [1.1K]3 years ago

7 0

Answer:

never D

Explanation:

I did it on edgenuity

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Which statement demonstrates that ultraviolet (UV) rays are electromagnetic waves?

MA_775_DIABLO [31]

The answer is B) They do NOT require a medium to travel.

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3 years ago

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nadezda [96]

Answer:

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3 years ago

An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi

Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0

3 years ago

Please help!!! I have a physics exam tomorrow and I just can't wrap my head around this one!

Ainat [17]

The total electrostatic force on charge A is $28 \mu N$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

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

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

$x_A = 0\\x_B = 10 m\\x_C = 20 m$

The magnitudes of the three charges are:

$q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C$

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

$F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N$

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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4 years ago

What happens if two small positively charged particles of equal magnitude are placed close to each other? A) The particles will

serious [3.7K]

That is not correct, it is C :)

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4 years ago

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