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garik1379 [7]
3 years ago
8

Beaker contain 200 mL of water what is volume of water in cm cube and m cube

Physics
2 answers:
Rainbow [258]3 years ago
7 0

1 mL = 1 cm³

1 cm³ = 0.000 001 m³

200 mL = 200 cm³

200 cm³ = 0.000 2 m³

It can be ANY substance. It doesn't have to be water.

maksim [4K]3 years ago
6 0

<u><em>Part 1:</em></u>

<u>The basics of conversions state that:</u>

1 ml is equivalent to 1 cm³

This means that the ratio between them is 1:1

Therefore, 200 ml of water would be equivalent to 200 cm³ of water

<u><em>Part 2:</em></u>

<u>Again, the basics of conversion states that:</u>

1 ml is equivalent to 1 * 10⁻⁶ m³

We will use cross multiplication to convert 200 ml to m³ as follows:

1 ml ................> 1 * 10⁻⁶ m³

200 ml .........> ??

200 ml = \frac{200*1*10^{-6}}{1}  = 0.0002 m³

Hope this helps :)

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Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

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