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Simora [160]
3 years ago
13

Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance

between the two objects were reduced to 5 m?
A. 1.5 N
B. 6 N
C. 12 N
D. 24 N
Physics
1 answer:
jeka943 years ago
8 0
Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N
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Explanation:

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3 years ago
What makes psychoanalysis different from behaviorism?
svp [43]
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4 0
3 years ago
A 6.00 cm tall light bulb is placed a distance of 54.2 cm from a double convex lens
BigorU [14]

Answer:

1 / f = 1 / i + 1 / o       thin lens equation

1 / i = 1 / f - 1 / o    =    (o - f) / (o * f)

i = o * f / (o - f)

i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm    image distance

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7 0
2 years ago
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
2 years ago
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algol [13]
Atomic Number
or
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5 0
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