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3 years ago

In a solution, the solvent is...

Physics

1 answer:

Anton [14]3 years ago

3 0

The solvent what the solute is dissolved in example: water

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A seesaw has an irregularly distributed mass of 50 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe

Simora [160]

Answer:

$d_{CM} = 0.78 m$

Explanation:

given,

mass of seesaw,m= 50 kg

length of seesaw, d = 3 m

mass of person on one end, m₁ = 60 Kg

mass of person on another end, m₂ = 86 Kg

position of center of mass = ?

System is in equilibrium

Clockwise moment is equal to anticlockwise moment.

$m_1 g\dfrac{d}{2} + m g d_{CM} = m_2 g \dfrac{d}{2}$

$m_1 d + 2 m d_{CM} = m_2 d$

$d_{CM} = \dfrac{m_2 d-m_1 d}{2 m}$

$d_{CM} = \dfrac{86\times 3 - 60\times 3}{2\times 50}$

$d_{CM} = 0.78 m$

hence, the position of center of mass from the fulcrum is 0.78 m on 60 Kg mass side.

4 0

3 years ago

Which two phases of the water cycle are repeated in this process? A. evaporation and precipitation B. evaporation and condensati

Katarina [22]

It Is A Evaporation And Precipitation

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3 years ago

The table shows the potential energy and kinetic energy of a skier at two different positions on a hill.

Serggg [28]

Answer:

30,000 units, because total energy remains unchanged

6 0

3 years ago

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At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig

Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

Second step: we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;

$y = ut - \frac{gt^2}{2}$

$y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}$

$y_{1} = -19.6m$

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1} = 5.4m[/tex]

Third step: we determine the y-coordinate of ball 2 after 2 seconds

$y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}$

$y_{2} = 10.4m$

Fourth step: we determine the y-component of the center mass of the two balls

$y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }$

$y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }$

y = 7.1m

Fifth step: we solve B part of the question; velocity of the center mass of the two balls

$Velocity = \frac{distance of center mass of the two balls (y)}{time}$

$velocity = \frac{7.1 m}{2 s}$

velocity = 3.55m/s

Sixth step: we solve C part of the question; acceleration of the center mass of the two balls

$acceleration = \frac{velocity}{time}$

$acceleration = \frac{3.55}{2}$

acceleration = 1.775 m/s^2

7 0

3 years ago

What three characteristics do geologist look for when observing a rock sample

lord [1]

Texture hardness and color

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3 years ago

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