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aleksklad [387]
3 years ago
13

On what number is the metric system based

Physics
1 answer:
Alexxx [7]3 years ago
3 0

The metric (or SI) system of units of measure is based on the number  10
and powers of it.

That's why people in Liberia, Burma, and the USA find it so confusing and
hard to learn, and why we stick to our familiar customary system based on
the number  3,  12,  1760,  5280,  16,  2,  4,  128, etc.

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A car speedometer has a 6% uncertainty. What is the range of possible speeds when it reads 100 km/h?
MAVERICK [17]

<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
</span>

3 0
3 years ago
What energy change is associated with the reaction to obtain one mole of H2 from one mole of water vapor? The balanced equation
Yakvenalex [24]

Answer:

ΔH = 249 kJ/mol

Explanation:

The balanced reaction is:

2H₂O(g) → 2H₂(g) + O₂(g)    (1)

To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:

H₂O(g) → H₂(g) + 1/2O₂(g)    (2)

The enthalpy of the reaction (2) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}  

<em>Where \Delta H_{r}: is the bond enthalpy of reactants and \Delta H_{p}: is the bond enthalpy of products.</em>

<u>For the reactants we have the next bond energies:</u>

2 x (H-O) = 2 x (467)

<u>And the bond energies for the products are:</u>

H-H + (1/2) (O=O) =  436 + (1/2)(498)

So, the enthalpy of the reaction (2) is:

\Delta H = 2 \cdot 467 kJ/mol - 436 kJ/mol - \frac{1}{2} \cdot 498 kJ/mol = 249 kJ/mol  

I hope it helps you!    

3 0
3 years ago
A 500 kilogram marlin and a 20 kilogram salmon are swimming at the same velocity. How does the momentum of the two fish compare?
ElenaW [278]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below are the choices that can be found from other sources:

A. The marlin has more momentum. 

<span>B. The salmon has more momentum. </span>

<span>C. Both have the same momentum. </span>

<span>D. Both have no momentum.
</span>
Velocity is speed. Momentum is the product of velocity and mass. Because the Marlin has a much greater mass that means that it's momentum would also be greater. 

<span>Lets look at this. momentum = mass x velocity </span>

<span>lets say that velocity for both fish is at 2 meters per second </span>
<span>Marlin : momentum = 500 x 2 = 1000 </span>
<span>Salmon : momentum = 20 x 2 = 40 </span>

<span>Therefore the answer is A</span>

6 0
3 years ago
Read 2 more answers
How much brighter is a sun-like star than the reflected light from a planet orbiting around it?.
Digiron [165]

It is about a billion times brighter.

<h3>What is the brightness of a star called?</h3>
  • A star's apparent magnitude, or how bright it seems to be from Earth, and absolute magnitude, or how brilliant it looks at a standard distance of 32.6 light-years, or 10 parsecs, are words used by astronomers to describe stellar brightness.
<h3>How bright are planets compared to stars?</h3>
  • Although planets and stars light up the night sky, planets usually appear much brighter than most stars.
  • Astronomers use astronomical scales to measure the relative brightness of celestial bodies, and many planets fall within the range easily visible to the eye.

To learn more about planets from the given link

brainly.com/question/1286910

#SPJ4

6 0
2 years ago
Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add
jok3333 [9.3K]

Answer:

When x = 2.8 cm, B_{x1} = 0.0265 T

When x = 5.5 cm, B_{x2} = 0.0209 T

when x = 7.3 cm, B_{x3} = 0.0169 T

When x = 11.0 cm, B_{x4} = 0.0103 T

Explanation:

According to Biot-Savart law,

B_{x} = \frac{N \mu_{o}IR^{2}  }{2(x^{2} +R^{2}  )^{3/2} }\\.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

\mu_{o}  = 4\pi  * 10^{-7} N/A^{2}

When x₁ = 2.8 cm = 0.028 m

B_{x1} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.028^{2} +0.11^{2}  )^{3/2} }\\B_{x1} = 0.0265 T

When x₂ = 5.5cm = 0.055 m

B_{x2} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.055^{2} +0.11^{2}  )^{3/2} }\\B_{x2} = 0.0209 T

When x₃ = 7.3 cm = 0.073 m

B_{x3} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.073^{2} +0.11^{2}  )^{3/2} }\\B_{x3} = 0.0169 T

When X₄ = 11.0 cm = 0.11 m

B_{x4} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.11^{2} +0.11^{2}  )^{3/2} }\\B_{x4} = 0.0103 T

4 0
4 years ago
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