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umka2103 [35]
2 years ago
6

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

e submerged. Determine the densities of:a. The glycerin b. The sphere.
Physics
1 answer:
Natasha_Volkova [10]2 years ago
6 0

Explanation:

weight of water displaced=weight of sphere,

Since,  plastic sphere floats in water with 50.0% of its volume submerged

so, water=2 sphere,

so sphere's density = 1/2 of water's  density

Now,  weight of glycerin displaced=weight of sphere,

Also given This same sphere floats in glycerin with 40.0% of its volume submerged.

so glycerin =2.5 sphere,

so sphere's density = .4 of glycerin's = 1/2 of water's.

So glycerin = 1.25 water

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PLEASE HELP ASAP!!
kolezko [41]

A. Wedge

and

B.Screw

A screw is like an inclined plane wrapped around a pole or rod.

and a wedge looks kind of looks like this / and is inclined.

Hope this helps!

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5 0
2 years ago
2. For each of the listed parts of a power plant, make a selection to indicate in what
Inga [223]

Answer: Find the answer in the explanation

Explanation: Given the Roman numeral and the representation

I. part of a coal-fired power plant

II. part of a nuclear power plant

III. part of a coal-fired power plant and part of a nuclear power plant

a.) Boiler : I

b.) Combustion chamber: I

c.) Condenser: I

d.) Control rod: II

e.) Generator: III

f.) Turbine: III

Toward the end processes part of both coal fire and nuclear power, they both make use of turbine and generator to generate electricity.

7 0
2 years ago
solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)
vodka [1.7K]
Using elimination, the answer is C. 
5 0
3 years ago
Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both
Nuetrik [128]

Answer:

Part a)

Width of the slit is

a = 580 nm

Part b)

Ratio of intensity is given as

\frac{I}{I_o} = 0.81

Explanation:

Part a)

As we know by the formula of diffraction we will have

a sin\theta = \lambda

so we have

\theta = 90

\lambda = 580 nm

so we will have

a sin90 = 580 nm

a = 580 nm

Part b)

As we know that the intensity in diffraction pattern is given as

I = I_o (\frac{sin\theta}{\theta})^2

\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2

so for angle 45 degree

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = 0.81

7 0
3 years ago
What is a conclusion?
Vikki [24]
The end or finish of an event or process.
8 0
2 years ago
Read 2 more answers
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