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3 years ago

Why can beta particles pass through materials more easily than alpha particles? (2 points)

Physics

2 answers:

never [62]3 years ago

5 0

Beta particles are smaller

Hunter-Best [27]3 years ago

3 0

Answer:

Beta particles are smaller.

Explanation:

When an unstable nuclei will break into stable nuclei then the nucleus energy of the unstable nuclei will be given to the the fast moving beta particles.

These beta particles are same as electron as they are negatively charged whose magnitude of charge is same as that of electrons charge.

so we have

$\beta^- = -1.6 \times 10^{-19} C$

now since electrons are smaller in size as we compare it with alpha particles so we can say that the beta particles can penetrate to more depth then the alpha particles

so here correct answer would be

Beta particles are smaller.

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2 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

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To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

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In the problem, the original centripetal acceleration is

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We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

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v' = 2v

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$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

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$r'=\frac{1}{4}r$

So the new acceleration is

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so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

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2 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

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By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$