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horrorfan [7]
3 years ago
13

The gravitational pull of Jupiter is greater than the gravitational pull of Earth. How would adding a “JUPITER MODE” to the virt

ual activity change the results of Galileo’s experiment?
Physics
1 answer:
Vlad1618 [11]3 years ago
5 0
The Galileo's experiment is about Earth’s gravitational force have on objects of different masses.
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Which of the following statements best describes the relationship between force and work?
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a).,  b).,  and  c).  are completely false. 
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In Physics, the technical definition of 'Work' is (force) times (distance).

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Which of the following statements about blood alcohol concentration (BAC) are true?
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D) Neither A or B are correct
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2 years ago
Read 2 more answers
Three conducting plates, each of area A, are connected as shown.
Shkiper50 [21]
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two. 
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂) 
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) ) 
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂ 
But I suppose we ought to kick that idea around a bit. 
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D. 
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁² 
Differentiate with respect to d₁ 
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero. 
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D 
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that 
d₁ = d₂ = ½D so 
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
7 0
3 years ago
Plz help will give brainlest! if you cant help, well, merry christmas!
soldi70 [24.7K]

most fossils are found in sedimentary rocks

8 0
2 years ago
The iron-rich core of the Moon is thought to be about how many kilometers in diameter?
natka813 [3]
The moon is thought to have an iron-rich core whose radius
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That puts its diameter in the range of  620 km to 700 km.
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