What do you mean by work?

Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

Learn more about gravitational constant:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

2 equal charges, 27 micro Coulomb each, are separated by 5 cm. Find force between those.

padilas [110]

Answer:

The force between charges is $F= 2.624*10^3N$ .

Explanation:

The Coulomb force $F$ between the two charges $q_1$ and $q_2$ separated by distance $d$ is given by the equation

$F = k\dfrac{q_1q_2}{d^2}$

where $k$ is the coulombs constant, and has the value

$k= 9*10^9N\cdot C\cdot m^2$ .

Now in our case

$q_1=q_2=27\mu C =27*10^{-6}C$

and

$d= 5cm =0.05m$ ,

therefore, the Coulomb force between the charges is

$F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}$

$\boxed{ F= 2.624*10^3N}$

3 0

4 years ago

The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at ti

chubhunter [2.5K]

Explanation:

Given that,

Initial speed of the rock, u = 30 m/s

The acceleration due to gravity at the surface of the moon is 1.62 m/s².

We need to find the time when the rock is ascending at a height of 180 m.

The rock is projected from the surface of the moon. The equation of motion in this case is given by :

$h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2$

It is a quadratic equation, after solving whose solution is given by:

t = 7.53 s

or

t = 8 seconds

(e)If it is decending, v = -20 m/s

Now t' is the time of descending. So,

$v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s$

Let h' is the height of the rock at this time. So,

$h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m$

or

h' = 155 m

6 0

3 years ago

When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed

bazaltina [42]

Answer:

$n_{fn}$ = 4

Explanation:

To solve this exercise we will use Bohr's atomic model

$E_{n}$ = - 13.606 / n² [eV]

The transition from level n = 2 to level n = 1 is valid

$E_{21}$ = - 13.606 [¼ -1/1]

$E_{21}$ = 10.2045 eV

Bohr's model for atoms with only one electron is

$E_{n}$ = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

ΔE = -13.606 [4 / $n_{fn}$ ² - 4/4]

4 / $n_{fn}$ ² = -ΔE / 13.606 + 1

4 / $n_{fn}$ ² = -10.2045 / 13.606 +1 = -0.75 +1

4 / $n_{fn}$ ² = 0.25

$n_{fn}$ = √ 4 / 0.25

$n_{fn}$ = 4

8 0

3 years ago

Science Seminar Question: Why did Vehicle 2 fall off the cliff in Claire's test of the collision scene but Vehicle 2 did not fal

asambeis [7]

Complete Question:

Check the file attached to get the complete question

Answer:

In the film Ice word Revenge, vehicle 2 did not fall of the cliff because, $Weight_{vehicle 1} < Weight_{vehicle 2}$ but in Claire's test, vehicle 2 off the cliff because $Weight_{vehicle 1} \geq Weight_{vehicle 2}$

Explanation:

In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff. In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.

8 0

3 years ago

What do you mean by work?​

What do you mean by work?