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3 years ago

Is that possible to separate the water particles from the air

Physics

2 answers:

Mashcka [7]3 years ago

6 0

Not likely but you can try

kupik [55]3 years ago

3 0

I don't really know but I don't think you can... I can't explain it either. Sorry im no help

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A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th

pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

Initial momentum of the returner, $P_i_1$ = 0 kgm/s
The initial momentum of the diving player, $P_i_2$ = 130 kgm/s

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

$P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s$

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0

2 years ago

What percentage does each parent contribute to a child's genotype?

bearhunter [10]

Each parent contribute 50 percent. A person have 46 genes and 23 comes from the mother and 23 comes from the father

5 0

3 years ago

Read 2 more answers

7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom

amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

$v_1t + v_2t = d$

$4*t + 28*t = 3.2 km$

$t = \frac{3.2}{32} = 0.1 hr$

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect

4 0

4 years ago

An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If

wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by 850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0

3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

0.85c

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = 0.85c

7 0

3 years ago