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Oliga [24]
3 years ago
9

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release

d from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released? a. As it moves farther and farther from Q, its speed will keep increasing.b. As it moves farther and farther from Q, its speed will decrease.c. As it moves farther and farther from Q, its acceleration will keep increasing.d. Its speed will be greatest just after it is released.e. Its acceleration is zero just after it is released.
Physics
1 answer:
UNO [17]3 years ago
6 0

Answer:

As it moves farther and farther away from Q, it's speed will keep increasing.

Explanation:

This is so because the point charges are both positive charges and positive charges repel.

As the charges repel, they do so at an increasing speed.

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A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity
Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

v = u + at

\therefore v=22+1.(6.8)

\Rightarrow v=22+10.88

\Rightarrow v=32.88 \ m/s

The final velocity become 32.88 m/s.

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2 years ago
Please help me asap ​
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Answer:

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8 0
3 years ago
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer
boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

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Frequency = speed ÷ wavelength
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