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4 years ago

In order to change the color of light you must change

Physics

1 answer:

kiruha [24]4 years ago

6 0

Answer:

The answer is C either the frequency or the wave lenght of the light

Explanation:

The frequency determins the brightnes of a color and the wave lenght determines the spectrum

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What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your

densk [106]

Answer:

The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

Explanation:

Given that,

Third charge $q_{3}=49.0\ nC$

Distance $x_{3}=-1.085\ m$

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

$F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}$

$F_{13}=-17268.3\times10^{-9}\ N$

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

$F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}$

Put the value into the formula

$F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}$

$F_{23}=11800.2\times10^{-9}\ N$

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

$F_{net}=F_{13}+F_{23}$

$F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}$

$F_{net}=-0.0000054681\ N$

$F_{net}=-5.468\times10^{-6}\ N$

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

5 0

3 years ago

The speed of a car is decreased uniformly from 30. meters per second to 10. meters per second in 4.0 seconds. What was the car's

masha68 [24]

Acceleration (magnitude anyway) = (change in speed) / (time for the change) .

Change in speed = (10 - 30) = -20 m/s

Time for the change = 4.0sec

Magnitude of acceleration = -20/4 = <em>-5 m/s² </em>

3 0

3 years ago

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Yanni just turned one. He loves to play with his cars and trucks. He can also name the animals he sees in books.

Liula [17]

Infancy
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3 years ago

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An object has a mass of 20kg, what is its weight on Earth?

umka2103 [35]

Answer: