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kondaur [170]
3 years ago
7

How many significant figures are there in : (a) 0.000054 (b) 3.001 x 10^5 (c) 5.600

Physics
1 answer:
melomori [17]3 years ago
4 0

Answer:

(a) 2 (b) 4 (c) 4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
  • All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
  • All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
  • All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

0.000054 has 2 significant figures.

3.001 x 10⁵ has 4 significant figures.

5.600 has 4 significant figures.

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Answer:

It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.

Explanation:

The linear coefficient of thermal expansion for steel is about 1.2*10^{-7}\°C^{-1}. From the equation of linear thermal expansion, we have:

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Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:

L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}

It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.

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3 years ago
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the
valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

T=2\pi\sqrt{\frac{m_c}{k}}     (1)

mc: mass of the chair

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T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

m_a=M-m_c=60.79kg-12.31kg=48.48kg

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BRAINLIEST IF CORRECT
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Hello There!

Sokka is here to help!!

The answer is...

<h2>D. Counter-arguments lead to circular logic in your argument.</h2>

Because, I am right. :)

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