Calculate the weight of a 4.5 kg rabbit.

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0

2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

Two students grab a slinky and start waving it up and down. A third student counts the number of waves that pass by every second

deff fn [24]

Velocity = frequency * wavelength

v = fλ, Just pick any points on the graph for frequency f and corresponding λ. Taking the first red point at the top. λ = 6m, f = 1 Hz, v = 6 * 1, v = 6 m/s

V = 6 M/S

4 0

2 years ago

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Which of the following best describes an action-reaction pair? A. The Moon Pulls on Earth, and Earth pulls back on the moon. B.

Papessa [141]

An action-reaction pair would be a pair in which one of the elements exerts a force on the other element (action), and then the other element would respond to this force by exerting another force in the opposite direction (reaction).

From the given choices, we will see that:
For choice A, the moon exerts a force on the earth by pulling it (action) and the earth responds to this force by pulling the moon (reaction in opposite direction of the action).

Therefore, the correct choice would be:
A. <span>The Moon Pulls on Earth, and Earth pulls back on the moon.</span>

4 0

2 years ago

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If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?

Vinvika [58]

The acceleration is0.25m/s^2

4 0

1 year ago

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