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Artemon [7]
1 year ago
10

IM A plastic sphere floats in water with 50.0% of its volume submerged. This same sphere floats in glycerin with 40.0 \% of its

volume submerged. Determine the densities of(b) the
Physics
1 answer:
arsen [322]1 year ago
6 0

Density of glycerin is 5/4 times water density.

Density of the sphere is 1/2 times water density.

<h3>How to find the density of sphere floats and glycerin?</h3>

Density is defined as the ratio of mass to volume, that is density is the amount of mass per unit of volume.

A substance's density exist defined as its mass per unit of volume. Density is most frequently represented by the symbol, however Latin letter D may also be utilized. If the mass or volume of an object is altered, the density of the thing may also change. Water, for example, has a specific density. An object will sink if its density exists greater than that of water; it will float if its density is lower.

Since a plastic sphere floats in water with 50% of its volume submerged.

So, the Density of the sphere = 1/2× water density

Since the same sphere floats in glycerin with 40% of its volume submerged.

So, the Density of the sphere = 4/100 × Glycerin density

= 2/5 × Glycerin density

Density of Glycerin = 5/2 of sphere density

Density of glycerin = 5/2 × 1/2 × water density

Density of glycerin = 5/4 × water density

To learn more about the volume of sphere floats refer to:

brainly.com/question/10827296

#SPJ4

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2) Charge

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k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C
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3 years ago
Merton’s strain theory would have the most trouble explaining which crime?
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A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6
mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87  = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

n=\dfrac{m}{M}

M=Molar mass

n=\dfrac{m}{M}

0.99=\dfrac{28.6}{M}

M=28.88 gm/mol

3 0
3 years ago
A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of
zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}

\dfrac{1}{f}=\dfrac{347}{5852}

f=16.86\ cm

We need to calculate the focal length

Using formula of magnification

m= \dfrac{-v}{u}

Put the value into the formula

2=\dfrac{v}{u}

v = -2u

Using formula of for focal length

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}

\dfrac{1}{16.86}=\dfrac{1}{2u}

2u=16.86

u=\dfrac{16.86}{2}

u=8.43\ cm

Hence, The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0
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