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exis [7]
2 years ago
7

How many significant digits are in the following measurements? a. 1300 m

Physics
1 answer:
Fofino [41]2 years ago
5 0

Answer:

For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).

Explanation:

hope it helps :)

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea
Vikentia [17]
For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c). 

fringe = (delta t) / (λ/c) 

We can find (delta t) with the equation: 

delta t = [v^2(L1+L2)]/c^3 

Derivation of this formula can be found in your physics text book. From here we find (delta t): 

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13 

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes 

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s. 
4 0
3 years ago
Cardiovascular exercise can
UkoKoshka [18]
Can <span>get your heart rate up and increases blood circulation throughout the body.</span>
3 0
3 years ago
Read 2 more answers
A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

d=2A

Where, d = distance

A = amplitude

Put the value into the formula

d=2\times0.08190

d=0.1638\ m

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

5 0
3 years ago
Which statement is true about effective nuclear charge?a. Effective nuclear charge decreases as we move to the right across a ro
MissTica

Answer:

Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table

Explanation:

The <em>effective nuclear charge </em>is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.

The <em>effective nuclear charge</em> is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.

The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.

Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.

The <em>effective nuclear charge</em>, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.

  • As the<em> effective nuclear charge</em> is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.

3 0
3 years ago
Anyone know what the answer is?..
Licemer1 [7]

Answer:

sometimes harmful and sometimes beneficial

8 0
3 years ago
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