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3 years ago

What are four elements on the periodic table

1 answer:

4 0

There are 92 natural ones and about 20 more made in particle-Physics labs.

If we only stick to the 92 natural ones, then there are

   (92!) / (88! times 4!) = 2,794,155 possible answers to your question.

Here's one: Osmium
      Neodymium
      Iridium
      Gold .

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Please hurry and please no links. I need an answer. Explain why a smart meter monitors the amount of electrical energy being use

brilliants [131]

Smart Metering is a technology encouraging end-use energy efficiency. Smart Meters can give consumers clear and comprehensive information about their energy consumption and, giving them better information, can help consumers to become more energy efficient.

Hope it helped you... pls mark brainliest

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3 years ago

A wave is incident on the surface of a mirror at an angle of 41° with the normal. what can you say about its angle of reflection

Umnica [9.8K]

It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

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3 years ago

Read 2 more answers

How many atoms are in SO<br> 3

kkurt [141]

Answer:

There are 6.87 x 1023 atoms in 1.14 mol SO3, or sulfur trioxide :

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3 years ago

Con base en el tema ley de ohm, resistividad y resistencia resuelva los siguientes

Bumek [7]

Answer:Explanation:gfgfgfgfgfgfgfgfgfg

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3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0

1 year ago