What quantity do units represent in a value 1. size 2.direction 3.dimension 4.magnitude

Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r

Alex787 [66]

Answer:

shut up nerd

Exption:

7 0

3 years ago

In 2020, nasa confirmed the existence of what on the lunar surface?.

kipiarov [429]

Water was confirmed to be on the sunlit surface of the Moon

6 0

2 years ago

What's the difference and similarity between O2 and 2 O ?

nasty-shy [4]

O2 refers to two oxygen atoms bonded together, while 2 O refers to two oxygen atoms that are not bonded to each other. They both have two oxygen atoms, but in O2, the oxygen atoms are bonded, while in 2O, the atoms are not.

4 0

3 years ago

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$ .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$ $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0

3 years ago

A lawnmower engine running for 20 minutes does 4560000 j of work. What is the power output of the engine

Alenkasestr [34]

Answer:

tim e (t) = 20 min.

= 20 × 60 = 1200 s ,

Work ( W) = 4560000 J

= 4560 KJ ,

Determine:

Power output (P) = Work ÷ time

= 4560 ÷ 1200

P = 3.8 KW

3 0

3 years ago

What quantity do units represent in a value1. size2.direction3.dimension4.magnitude