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My name is Ann [436]
3 years ago
6

The unit of length most suitable for measuring the thickness of a cell phone is a ( megameter, meter, millimeter, nanometer ) Th

e unit of length most suitable for measuring the height of a backyard tree is a ( megameter, meter, millimeter, nanometer ).
Physics
1 answer:
puteri [66]3 years ago
5 0

Answer: 1.millimeter 2.meter

Explanation:

a phones thickness can be measured with such a small measurement and a tree is not big enough for a megameter but not small enough for a millimeter so it'd be a meter

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A 10N force and a 50N force act in opposite directions on a 4 kg object. What is the acceleration? ( how did you do it )
Anastaziya [24]

To find the acceleration, we will use the formula

F=ma

Here the net force will be 40N (50 - 10)

F = ma

40 = 4a

40/4 = a

10 m/s^2 = a

Hope you understood !

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1. Most of the stars on the HR Diagram are classified as which type of star?
xenn [34]

Answer:

1. Main Sequence - middle life 17

2. red

3. blue

4. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16.

5. red giants

Explanation:

Main sequence stars have a Morgan-Keenan luminosity class labeled V. red giant and supergiant stars (luminosity classes I through III) occupy the region above the main sequence. They have low surface temperatures and high luminosities which, according to the Stefan-Boltzmann law, means they also have large radii. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16. The hottest stars are the blue stars. A star appears blue once its surface temperature gets above 10,000 Kelvin, or so, a star will appear blue to our eyes. The lowest temperature stars are red while the hottest stars are blue. Astronomers are able to measure the temperatures of the surfaces of stars by comparing their spectra to the spectrum of a black body. Most stars, including the sun, are "main sequence stars," fueled by nuclear fusion converting hydrogen into helium. ...

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3 years ago
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A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3
Zinaida [17]

Answer:

a. \bar{d}=4.34 cm

b. \sigma=0.023 cm

c. \rho=(0.0089\pm 0.00058) kg/cm^{3}

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}

  • x_{i} is values of each diameter
  • N is the total number of values. N=6

\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}

\bar{d}=4.34 cm

b) The standard deviation equations is:

\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}

If we put all this values in that equation we will get:

\sigma=0.023 cm

Then the mean diameter will be:

\bar{d}=(4.34\pm 0.023)cm

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: V=\pi R^{2}h

Then:

\rho=\frac{m}{\pi R^{2}h}

Using the values that we have, we can calculate the value of density:

\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}

We need to use propagation of error to find the error of the density.

\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}  

  • δm is the error of the mass value.
  • δd is the error of the diameter value.
  • δh is the error of the length value.

Let's find each partial derivative:

1. \frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089

2.  \frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004

3. \frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071

Therefore:

\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}

\delta\rho=0.00058

So the density is:

\rho=(0.0089\pm 0.00058) kg/cm^{3}

I hope it helps you!

3 0
3 years ago
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