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andrezito [222]
3 years ago
5

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

roton and an electron are situated 911 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
Physics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 911 nm = 911 x 10^-9 m

The Coulomb's force is given by

F=\frac{Kq_{1}q_{2}}{r^{2}}

F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9}  \right )^2}

F = 2.78 x 10^-16 N

The force between the electron and the proton is 2.78 x 10^-16 N.

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CAN YOU PLS CHECK IF ITS CORRECT I'LL MARK YOU BRAINLIST
Alexus [3.1K]

i think it looks good

yea it correct

BTW yw if it's right

5 0
3 years ago
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A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t
UkoKoshka [18]

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

v_y^2 = u^2 + 2 g h

v_y^2 = 0^2 + 2 g h

v_y = \sqrt{2gh}

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v_y = \sqrt(211.68)

v_y = 14.55 m/s

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8 0
3 years ago
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position. Determine the
Eva8 [605]

Answer:

The frequency does not depend on the amplitude for any (ideal) mechanical or electromagnetic waves.

In electromagnetism we have that the relation is:

Velocity = wavelenght*frequency.

So the amplitude of the wave does not have any effect here.

For a mechanical system like an harmonic oscillator (that can be used to describe almost any oscillating system), we have that the frequency is:  

f = (1/2*pi)*√(k/m)

Where m is the mass and k is the constant of the spring, again, you can see that the frequency only depends on the physical properties of the system, and no in how much you displace it from the equilibrium position.

This happens because as more you displace the mass from the equilibrium position, more will be the force acting on the mass, so while the "path" that the mass has to travel is bigger, the mas moves faster, so the frequency remains unaffected.

5 0
3 years ago
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The distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the gr
Harman [31]

If the distance is proportional to the square of the time, then the distance in 9 seconds will be  (9/8)²  as long as the distance in 8 seconds.

(9/8)² = 1.265625

(64 ft) · (1.265625) = <em>81 ft. </em>

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4 years ago
The moon rotates on its axis relative to the distant stars. true or false
Vinvika [58]

Answer:

true

Explanation:

7 0
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