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AURORKA [14]
4 years ago
8

What is the density of a 36 g object with a volume of 15 cm3? (Density: D = )

Physics
2 answers:
Akimi4 [234]4 years ago
8 0
D = <span>2.4 g/cm3

We learned this in my physics class as well. 

</span><span><span>p = m/v</span></span>

Where:
p = density
m = mass
V = volume

So to find the density you solve 36/16 

=2.4

Hope this helped. Have a great day!

oksian1 [2.3K]4 years ago
8 0
Here, We know, Density = Mass / Volume
Here, mass = 36 g
volume = 15 cm³

Substitute their values, 
d = 36 / 15
d = 2.4 g/cm³

In short, Your Answer would be 2.4 g/cm³

Hope this helps!
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Convert: 8 mm: ____________ cm
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7 0
3 years ago
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
4 years ago
Read 2 more answers
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