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Sveta_85 [38]
3 years ago
11

As an object sinks in a fluid, the buoyant force ____.

Physics
2 answers:
tigry1 [53]3 years ago
6 0

As an object sinks in a fluid, the buoyant force decrease.



Hope this helped.


kiruha [24]3 years ago
3 0
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.

As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.

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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
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Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

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valentinak56 [21]

Answer:

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Explanation:

efficiency =   \frac{useful \: work \: done}{total \: energy \: input}

800 / 1000 = 0.8

Efficiency = 0.8 *100 = 80%

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