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GREYUIT [131]
3 years ago
12

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

ric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of −3.45×10⁻³ V . The charge and the mass of an alpha particle are α = 3.20×10⁻¹⁹ C and mα = 6.68×10⁻²⁷ kg , respectively.
Physics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

Speed of the alpha particle is v=1.8180\times 10^3m/sec      

Explanation:

We have given charge on alpha particle q=3.2\times 10^{-19}C

Mass of the alpha particle m=6.68\times 10^{-27}kg

Potential difference V=-3.45\times 10^{-3}volt

We have to find the speed of the alpha particle

From energy conservation we know that

\frac{1}{2}mv^2=qV

\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}

v=1.8180\times 10^3m/sec

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Answer:

Option (D) : The object slows down.

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3 years ago
A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time
baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

i = \frac{q}{t}

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

i = increased

So correct answer will be

B. The current increases.

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3 years ago
Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

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Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required  = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0
3 years ago
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What we can see from the formula is that, since the 2\pi does not change its value, the angular speed depends only on the period T.

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Thus, since the period for both is the same, the angular speed given by

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11. Which of the following is a proper unit of
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D is the correct answer!!
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