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Nastasia [14]
3 years ago
15

AWARDING BRAINLIEST (~ ̄▽ ̄)~

Physics
1 answer:
monitta3 years ago
3 0
False
Just took this test
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In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir
ale4655 [162]

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

5 0
3 years ago
What is the resultant force acting on an object? The first force is 45 N West and the second is 23 N East.
Paul [167]

Answer:

22N West

Explanation:

45-23 because they are in opposite directions.

4 0
3 years ago
A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?
Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book  = 1.15kg

Unknown:

Magnitude of the normal force  = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

 Normal force  = mass x acceleration due to gravity

 Normal force  = 1.15 x 9.8  = 11.27N

5 0
3 years ago
If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi
Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

I\alpha \frac{1}{r^{2} }

Now according to the question the distance is increased by three times than,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }

Therefore,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }

Therefore the intensity will become 1/9 times to the initial intensity.

3 0
3 years ago
What is the energy of an electromagnetic wave that has a frequency of
sukhopar [10]

Answer:

Energy, \; E = 2.6504 * 10^{-34} \; Joules

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34}

Energy, \; E = 2.6504 * 10^{-34} \; Joules

8 0
3 years ago
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