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Citrus2011 [14]
3 years ago
12

If we have two objects with the same mass but different densities (Lets assume object 1 is denser,therefore volume is lower rela

tive to object 2).Assume both objects have a lower density than water.Now if we put both in water,both will float.Since they float, buoynacy force is equal to the weight of the object,so buoynacy force in 1 and 2 should be equal .However if we use F(boyancy)=pvg--> then becuase the submerged v is less for the denser object and p fluid is the same,then we can conclude that F bouynacy are not the same for both objects. True or false?

Physics
1 answer:
o-na [289]3 years ago
7 0

take the volume that is submerged. That's the volume of water displaced.

but the volume of submerged is different in those two

so buoyancy force is different in those two

weights are the same

probably. since the densities* are different.

In question it says wights are the same and diffferent volumes

so it seems that the one with more density should have a lower buyonacy force

but you said that when an object floats the buoyancy force equals weight,so since both objects have the same weight,then buoynacy force should be equal in those two


The more dense object will float with a greater percentage of its volume immersed, not less.

2) If they have the same MASS, the more dense one will have less VOLUME

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What is the strength of the electric field in a region where the electric potential is constant?
Roman55 [17]

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx        ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

<em>Therefore, a constant electric potential means that electric field is zero.</em>

4 0
3 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

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distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
3 years ago
A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If
Damm [24]

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Rearrange the equation:

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We also know the capacitance of a parallel-plate capacitor is given by:

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κ is the capacitor's dielectric constant

ε₀ is the electric constant

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If we substitute C:

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We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

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Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
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The solution to the problem is as follows:

 <span>Average = 80 
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What is a force field
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Chiefly in science fiction) an invisible barrier of exerted strength or impetus.
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