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anastassius [24]
3 years ago
8

A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.

Which one has the greater buoyant force on it? Please explain. Answer: B - the aluminum
A) Both have the same buoyant force.
B) the aluminum
C) the lead
D) It cannot be determined without knowing their volumes.
Physics
2 answers:
AysviL [449]3 years ago
8 0

The buoyant force on a submerged object is the weight of the water it displaces ... the water it pushes out of the way.  That amount is simply the volume of the submerged object.  So the more volume is submerged, the greater will be the buoyant force acting on it.

Since Aluminum is less-dense than lead, the same 10kg of Aluminum needs a bigger container to hold it than 10kg of lead needs.  The aluminum needs  more volume to hold the same mass.

The aluminum displaces more water.  So the buoyant force acting on the <em>aluminum</em> is greater than the buoyant force acting on the lead.<em> (B) </em>.

I'm guessing this is a big part of the reason why fishing sinkers are not made of aluminum.

Lelechka [254]3 years ago
5 0
I think that the answer is B
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A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged unifor
Zigmanuir [339]

Answer:

Qsinθ/4πε₀R²θ

Explanation:

Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.

Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.

So dq = λRdθ.

Substituting dq into dE', we have

dE' = dqcosθ/4πε₀R²

= λRdθcosθ/4πε₀R²

= λdθcosθ/4πε₀R

E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ

E' = (λ/4πε₀R)[sinθ] from -θ to θ

E' = (λ/4πε₀R)[sinθ]

= (λ/4πε₀R)[sinθ - sin(-θ)]

= (λ/4πε₀R)[sinθ + sinθ]

= 2(λ/4πε₀R)sinθ

= (λ/2πε₀R)sinθ

Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ

Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ

Q = 2λRθ

λ = Q/2Rθ

Substituting λ into E', we have

E' = (Q/2Rθ/2πε₀R)sinθ

E' = (Q/θ4πε₀R²)sinθ

E' = Qsinθ/4πε₀R²θ where θ is in radians

 

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3 years ago
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If we knew the car mass, we could find the hill slope angle.

If we knew the hill slope angle, we could find the car mass.

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cupoosta [38]

Answer:

Read below!

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An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.  

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4 0
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I believe they would electron rate would slow down and the molecules would shrink.

I am almost positive that this is correct. I hope it helps!

6 0
3 years ago
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