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Aliun [14]
3 years ago
15

According to Newton's third law of motion, if you push against a wall, the wall will __________.

Physics
2 answers:
slavikrds [6]3 years ago
3 0

Answer:

push up agaisnt you with equal force.

egoroff_w [7]3 years ago
3 0

Answer:

c

Explanation:

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Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less
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Are outer planets gaseous
bija089 [108]

Answer:

Yes

Explanation:

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3 years ago
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe
deff fn [24]

Answer:

\theta_2 - \theta_1 = 156.93 degree

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

y = A sin\omega t

y = \frac{A}{5}

so now we have

\frac{A}{5} = A sin\omega t

now we have

\theta_1 = 11.53 degree

so the phase other particle in opposite direction is given as

\theta_2 = 180 - 11.53 = 168.46 degree

so we have phase difference given as

\theta_2 - \theta_1 = 168.46 - 11.53

\theta_2 - \theta_1 = 156.93 degree

7 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:

<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
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