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oee [108]
3 years ago
5

A positively charged sphere with a charge of the of 8Q is separated from a negatively charged sphere - 2Q by a distance r. The s

pheres briefly touch each other and move to the original distance are. what is the new force on each sphere inn terms of F
Physics
1 answer:
professor190 [17]3 years ago
4 0

Answer: The new force is:

F1 = -(9/16)*F0

where F0 is the initial force.

Explanation:

The charges of the spheres is q1 = 8Q and q2 = -2Q

and as you know, the force between charged objects is:

F = k*q1*q2/r^2

where k is a constant and r is the distance

F0 = -(k16Q^2)/r^2

where the negative sign means that the force is attractive.

Now, when the spheres touch each other, the charge must be distributed equally in both spheres. So the total charge is q1 + q2 = 8Q - 2Q = 6Q

then each sphere has now a charge of 3Q.

The new force is:

F1 = k*3Q*3Q/r^2 = (k*9*Q^2)/r^2 = -(9/16)*F0

You can see that F1 is positive, this means that the force now is repulsive.

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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
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Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

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3 years ago
Which is an example of observational learning?
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The action that has the ability to change the motion of an object is
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3 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

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Substitute the known values into the equation.

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The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

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