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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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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
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Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

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