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Novay_Z [31]
3 years ago
9

In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir

m and a very few bounced back. Suppose instead that about one-half the alpha particles bounced back and one-half went through. How would this have changed his conclusion about the structure of the atom?
1. The nucleus is about 1/2 the size of the atom.
2. The atom has no nucleus.
3. None of these
4. The nucleus covers the entire atom.
5. The atom is mostly empty space.
Physics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

1. The nucleus is about 1/2 the size of the atom

Explanation:

Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size

Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles

so correct answer will be

1. The nucleus is about 1/2 the size of the atom

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2 years ago
Why echo sound cannot be heard in small room
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They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
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3 years ago
FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine
Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0
1 year ago
A fold is a _ in a rock , and a fault is a _in a rock?
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So A fold is a Bend? in a rock. Maybe.

</span>A fault<span> is a planar fracture or discontinuity in a volume of </span>rock<span>, across which there has been significant displacement as a result of </span>rock<span>-mass movement.</span>
3 0
3 years ago
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Answer:

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