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Kitty [74]
3 years ago
13

What volume In liters is a cube 20cm on a side? If the cube is filled with water what is the mass of the water?

Physics
1 answer:
frez [133]3 years ago
7 0
1 liter = 1000 cm^3
20cm * 20cm * 20cm = 8000 cm^3
8000/1000 = 8 liters

Since 1ml of water = 1 cm^3 = 1 grams
8 liters = 8000 grams = 8 kilograms
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Two children of mass 18 kg and 29 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c
DedPeter [7]

Answer:

3.085 [m].

Explanation:

1) The rule:

m₁*g*l₁=m₂*g*l₂, where m₁ and l₁ - the mass and distance for the small child, m₂ and l₂ - for the big child;

2) according to the condigion l₁+l₂=5, then

3) it is possible to make up the system:

\left \{ {{l_1+l_2=5} \atop {m_1*l_1=m_2*l_2}} \right. \ = > \ \left \{ {{l_1=5-l_2} \atop {18*(5-l_2)=29*l_2}} \right. \ = > \ \left \{ {{l_1=\frac{145}{47}} \atop {l_2=\frac{90}{47}}} \right.

4) finally, l₁=145/47≈3.085 [m].

7 0
2 years ago
Light traveling through plate glass (n = 1.52) approaches the surface, i.e. the air-glass interface. What minimum incident angle
Ilia_Sergeevich [38]

Answer:

41°

Explanation:

refractive index of glass, n = 1.52

For total internal reflection, the angle of refraction in rarer medium is 90°.

r = 90°

Let the angle of refraction in denser medium, that means in glass is i.

By use of Snell's law

refractive  index of rarer medium with respect to denser medium, that means refractive index of air with respect to glass = 1 / n

And

\frac{1}{n}=\frac{Sin i}{Sin r}

\frac{1}{1.52}=\frac{Sin i}{Sin 90}

0.658= Sin i

i = 41°

Thus, the angle of incidence should be 41°.

8 0
3 years ago
Which formation moves sediment and erodes rock?
borishaifa [10]

Answer:

C

Explanation:

7 0
3 years ago
Read 2 more answers
Durning which type of process does pressure remain consistent
Simora [160]

Answer:

\fbox {D. Isobaric}

Explanation:

The process during which pressure remains constant is called an isobaric process.

6 0
2 years ago
Read 2 more answers
If this did not happen, what would be the approximate force on an eardrum of area 0.22 cm2 if a change in altitude of 1500 m tak
jeyben [28]

Complete Question

When you ascend or descend a great deal when driving in a car yours ears "pop," which means that the pressure behind the eardrum is being equalized to that outside. If this did not happen, what would be the approximate force on an eardrum of area .50 cm2 if a change in altitude of 950 m takes place?

Answer:

The value is F    = 0.60 \  N

Explanation:

From the question we are told that

   The area of the ear drum is  A = 0.5 \  cm^2 = 0.50 *10^{-4} \  m^2

    The change in altitude is  \Delta d  = 950 \  m

Generally the change in pressure is mathematically represented as

       \Delta P = \frac{F}{A}

This can also be mathematically represented as

      \Delta P = \rho * g *  \Delta d

So

       \frac{F}{A}     = \rho * g *  \Delta d

=>    F    = \rho * g *  \Delta d  *  A

=>    F    = \rho * g *  \Delta d  *  A

Here \rho is the density of dry air with value  \rho =  1.29 \ kg /m^3

So

     F    = 1.29 * 9.8 *  950  *  0.50 *10^{-4}

=> F    = 0.60 \  N

3 0
3 years ago
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