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liubo4ka [24]
3 years ago
10

25 POINTS In outerspace, do Newton's laws still apply? A) yes B) No

Physics
2 answers:
Alla [95]3 years ago
6 0
Maybe is “yes” I guess
Nady [450]3 years ago
6 0

Answer:

No

Explanation:

There is no gravity

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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:

<em>0.85c </em>

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 = γE_{0}

which can be written in this case as

E_{K} = γE_{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 = <em>0.85c </em>

7 0
3 years ago
What is this Regeneration​
Mila [183]

Answer:

Regeneration means that an organism regrows a lost part, so that the original function is restored.

5 0
3 years ago
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Within which type of water body does water move the slowest?
lawyer [7]
It's D i'm pretty sure
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4 years ago
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Help plzzzz thank youu
dybincka [34]
The answer is B I hope this helps luv
5 0
4 years ago
The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^
Arlecino [84]

The solution for this problem is:

 
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s² 
this is for t in seconds and r in meters 

v = dr/dt = [0.180t i - 0.0450t² j] m/s² 

tan(-36.0º) = -0.0450t² / 0.180t 
0.7265 = 0.25t 
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7 0
3 years ago
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