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3 years ago

How do you know the speed of an electromagnetic wave in a vacuum?

1 answer:

5 0

Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.

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I HAVE 5 MINUTES!!!!!! A block oscillating on the end of a spring moves from its position of maximum spring stretch to maximum s

Verdich [7]

Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.

4 0

3 years ago

Explain any 3 alternating ways of producing electricity

ehidna [41]

Answer:

to produce electricty three ways you have the water wheel the wind mill and solor panels

Explanation:

4 0

3 years ago

Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o

Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

$F_{23} = \frac{k*q_{2}*q_{3}}{x^2}$ (Electric force of q₂ over q₃)

$F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

$\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (We cancel k and q₃)

$\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}$

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0

3 years ago

Read 2 more answers

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m$

The displacement of the ball on the moon is 7.04 m

6 0

4 years ago

What affects the vertical velocity of a projectile over time?<br>---

kobusy [5.1K]

The vertical velocity is affected by the acceleration of gravity (ignoring the effects of air resistance usually)

6 0

3 years ago