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3 years ago

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

he proton must impact the nucleus with a kinetic energy of 1.80 MeV. Assume the nucleus remains at rest.
Required:
a. With what speed must the proton be fired toward the target?
b. Through what potential difference must the proton be accelerated from rest to acquire this speed?

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

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I believe this is what you have to do:

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Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

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As well as the Normal Force $N$ :

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In addition, we know $N=w$ , then $\sum F_{y}=0$

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Substituting (1) in (5):

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In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

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On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

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We already know the value of $d$ and calculated $a$ , we have to find $t$ :

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3 years ago