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lubasha [3.4K]
3 years ago
5

What is the amount of solvent in an amount of solute?

Physics
1 answer:
jarptica [38.1K]3 years ago
3 0

Answer:

The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.

Explanation:

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Which principle allows us to conclude that gravity acts as the same today and will tomorrow?
Svet_ta [14]

Answer:

Gravity is a constant quantity

Explanation:

4 0
3 years ago
Q11:A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5
natulia [17]

Answer:

   v_s = 34.269 m / s

Explanation:

This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.

         f ’= f  \frac{v}{v  \mp v_s  }

where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer

In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together

          5500 = f (\frac{v}{v - v_s })

          4500 = f ( \frac{v}{v + vs})

let's solve these two equations

           \frac{5500}{4500}     = \frac{v+v_s}{v-v_s}

           1.222 (v-v_s) = v + v_s

            v_s (1+ 1.22) = v (1.222 -1)

            v_s = v   \frac{0.222}{2.223}

the speed of sound in air is v = 343 m / s

            v_s = 343 0.09990

 

             v_s = 34.269 m / s

4 0
3 years ago
When one form of energy is converted into another what is the name of that
Svetradugi [14.3K]
Transformation of energy, because energy cannot be created or destroyed
8 0
3 years ago
a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance
8_murik_8 [283]

The car's (average) acceleration would be

a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}

The car's position over time would be given by

x=v_0t+\dfrac12at^2

so that after 2.4 seconds, the car will have traveled a distance of

x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2

\implies x=77.5\,\mathrm m

7 0
3 years ago
Read 2 more answers
Now imagine a person dragging a 50 kg box along the ground with a rope, as
ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

                                                     = 50 x 9.8

                                                     = 490 N

The coefficient of friction,

                            μ = f/ η

                               = 30 / 490

                               = 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
4 years ago
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