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1 year ago

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what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

ectron that is accelerated from rest through a potential difference of 1000 v?

1 answer:

Evgen [1.6K]1 year ago

5 0

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

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A current runs from the left to the right in a long, straight wire. How does the magnetic field at point X compare with the magn

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I think the answer is <span>D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.</span>

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3 years ago

Why might a scientist repeat an experiment if she did not make a mistake in the first one?

iren2701 [21]

Answer:

B

Explanation:

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3 years ago

Two balloons are charged with an identical quantity and type of charge: -6.25 x 10-6 C. They are held apart at a separation dist

steposvetlana [31]

Answer:

The electric force between them is 878.9 N

Explanation:

Given:

Identical charge $q = -6.25 \times 10^{-6}$ C

Separation between two charges $r = 0.02$ m

For finding the electrical force,

According to the coulomb's law

$F = \frac{k q^{2} }{r^{2} }$

Here, force between two balloons are repulsive because both charges are same.

Where $k = 9 \times 10^{9}$

$F = \frac{9 \times 10^{9} \times (-6.25 \times 10^{-6} )^{2} }{4 \times 10^{-4} }$

$F = 878.9$ N

Therefore, the electric force between them is 878.9 N

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3 years ago

What is the collective noun of chocolate

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3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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2 years ago