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Veseljchak [2.6K]
2 years ago
6

A 2.0-mm-diameter glass sphere has a charge of 1.0 nC. What speed does an electron need to orbit the sphere 1.0 mm above the sur

face
Physics
1 answer:
Lorico [155]2 years ago
3 0

The electron would need to attain a speed of at least 2.813*10^7 m/s

Data;

  • diameter of sphere = 2.0mm
  • charge = 1.0 nC

The distance of the electron from the center of the sphere  = (1 + 1)mm = 2mm.

2mm = 2.0*10^-3m

<h3>Centrifugal Force and Coulomb's law</h3>

Using centrifugal force of attraction and coulomb's law, we can determine the speed which the electron needs.

\frac{mV^2}{R}=\frac{Kq_1q_2}{r^2}\\ K = \frac{1}{4\pi \epsilon } \\ V^2 = \frac{q_1R\\}{4\pi \epsilon \ m r} \\v = \sqrt{\frac{1.0*10^-^9*1.602*10^-^1^9}{4\pi \epsilon *9.1*10^-^3^1*2.0*10^_3} } \\

Solving the above, we would have

v = 2.813*10^7m/s

The electron would need to attain a speed of at least 2.813*10^7 m/s

Learn more on centrifugal and coulomb's law here;

brainly.com/question/24743340

brainly.com/question/10220774

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Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
2 years ago
Kinetic Energy - What does it depend on?
Greeley [361]

Answer:

faster; more kinetic energy

Explanation:

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2 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
2 years ago
A jet plane comes in for a downward dive as shown in the figure below.
xxTIMURxx [149]

The solution would be like this for this specific problem:

<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g = v^2/r </span>
<span>v^2 = 4.5 g * r </span>
<span>v = sqrt ( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124 m/s</span>

So the pilot will black out for this dive at 124 m/s. I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

8 0
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Waves diffract the most when their wavelength is
djverab [1.8K]
<h2>Answer: about the same size of the gap  or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

<u />

5 0
2 years ago
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