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-Dominant- [34]
3 years ago
8

A student has a sample of an insoluble solid and a liquid. The task is to separate the mixture into its pure components. What qu

estion is best asked in the design of the experiment to determine how to separate the components?
A) Is the solid/liquid mixture toxic in its combined form?

B) Is the solid large enough to be caught by the pores of a filter?

C) Is the melting temperature of the solid lower than that of the liquid?

D) Can the solid be crushed into smaller particles to allow for distillation?
Physics
1 answer:
neonofarm [45]3 years ago
6 0

Answer:

<em><u></u></em>

  • <em><u>B) Is the solid large enough to be caught by the pores of a filter? </u></em>

Explanation:

An <em>insoluble solid</em> and a <em>liquid</em> form an heterogeneous mixture, meaning the liquid and the solid could be <em>separated</em> by physical media.

If the size of the solid particles are large enough it could be separated from the liquid by filtration.

Since, the task is to separate the mixture into its pure components, the best and first step is to determine whether the solid particles have the size that permit to retain them in the available fliter, which is described by<em> </em>the option <em>B): is the solid large enough to be caught by the pores of a filter? </em>

As for the other options:

<em>A) Is the solid/liquid mixture toxic in its combined form?</em>

  • a mixture is not more toxic than its pure components, because they are a physical combination and not a chemical one.

<em>C) Is the melting temperature of the solid lower than that of the liquid?</em>

  • you do not want to melt the solid, because that would make the separation more difficult.

<em>D) Can the solid be crushed into smaller particles to allow for distillation? </em>

  • crushing the solid will not make it soluble and will not change the boiling point of the solution, thus this is not a step to separate the mixture into its components.
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raketka [301]

Answer:

iv) It is 9x bigger than before

Explanation:

As the amplitudes of the new speakers add directly with the original one, taking into account the phase that they have, the composed amplitude of the sound wave is as follows:

At = A + 4A -2A = 3 A

The intensity of the wave, assuming it propagates evenly in all directions, is constant at a given distance from the source, and can be expressed as follows:

I = P/A

where P= Power of the wave source, A= Area (for a point source, is equal to the surface area of a sphere of radius r, where is r is the distance to the source along a straight line)

For a sinusoidal wave, the power is proportional to the square of the amplitude, so the intensity is proportional to the square of the amplitude also.

If the amplitude changes increasing three times, the change in intensity will be proportional to the square of the change in amplitude, i.e., it will be 9 times bigger.

So, the statement iv) is the right one.

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3 years ago
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
Select ALL of the following statements that provide evidence that there is friction acting on a cart moving along a level track.
Sati [7]

Answer:

1st and 4th one................

6 0
3 years ago
Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli
Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = v_{f}=?

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}

Substituting the values we get

2\times 14 + 5\times 0 =(2+5)\times v_{f}

v_{f}=\dfrac{28}{7}=4\ m/s

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

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3 years ago
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strojnjashka [21]

the correct answer is D i just got it on USATestPrep. Your welcome.

7 0
3 years ago
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