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KIM [24]
3 years ago
11

An example of a solution in which the solute is not a solid and the solvent is not a liquid.

Physics
1 answer:
sashaice [31]3 years ago
6 0
The solute can be acetic acid and solvent can be water, which gives us dilute acetic acid.
We can say 2 gases as both a solvent and solute, such as nitrogen gas and oxygen gas. (Either one can be the solvent or solute)
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You are sitting in a car that is travelling at a constant 10 m/s. How fast is your drink, sitting in the cup holder, traveling r
Ksenya-84 [330]

Here, we are required to determine how fast is you drink, sitting in the cup holder, travelling relative to the car.

  • The speed of the drink, sitting in the cup holder, relative to the car is; 0m/s

From the laws of relative motion,

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)</em>

  • <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)when object A and Object B are travelling with speed a and b respectively in the same direction, where speed a = speed b, then the speed of object A relative to object B is; zero(0).</em>

Evidently, the scenario in the question is similar to the third scenario above. The cup, sitting in the cup holder is travelling with the car at the same constant speed 10m/s.

Therefore, the speed of the drink relative to the car is zero(0).

Read more:

brainly.com/question/20549055

7 0
2 years ago
Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the
Leni [432]

Answer:

Part(a):  The relative capacitance is \dfrac{C_{A}}{C_{B}} = 0.33

Part(b): The relative energy stored is \dfrac{U_{A}}{U_{B}} = 0.33

Part(c): The relative charge stored is \dfrac{Q_{A}}{Q_{B}} = 0.33

Explanation:

We know the capacitance (C) of a capacitor having charge (Q) and subjected to a potential difference of (V) is given by

C = \dfrac{Q}{V}

Also, the energy (U) stored by a capacitor can be written as

U = \dfrac{1}{2}C~V^{2}

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be \textbf{r_{i}^{B}}\bf{r_{i}^{B}}, the outer radius be \bf{r_{o}^{B}}, the inner radius of Capacitor A be \bf{r_{i}^{A}} and the outer radius be \bf{r_{o}^{B}}.

Given in the problem,

&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}

Now, the capacitance (C) of a cylindrical capacitor is given by,

\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}

where \epsilon_{o} is the permittivity of the free space, L is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}

and the capacitance of capacitor B,

C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}

giving the relative capacitance of each capacitor to be

\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33

Part(b):

Energy stored by capacitor A,

U_{A} = \dfrac{1}{2}~C_{A}~V^{2}

Energy stored by capacitor B,

U_{B} = \dfrac{1}{2}~C_{B}~V^{2}

giving the relative energy stored by each capacitor to be

\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33

Part(c):

The charge stored by capacitor A,

Q_{A} = C_{A}~V

The charge stored by capacitor B,

Q_{B} = C_{B}~V

giving the relative charge stored by each capacitor to be

\dfrac{Q_{A}}{Q_{B}} =  \dfrac{C_{A}}{C_{B}} = 0.33

8 0
3 years ago
Rowenda is stopped on the side of the road. She then starts pedaling her bicycle for 15 seconds. Her final
lukranit [14]

Explanation:

acceleration = change in velocity / change in time

a = v2-v1

- - - - - -

t

V1 = 0 ( since she stopped)

V2 = 9 m/s

t =15s

a = 9 - 0

- - - -

15

= 0.6m/s^2

5 0
3 years ago
Which of the following is a resistor?<br> fan<br> switch<br> battery<br><br><br> Thank you!
Rama09 [41]
The answer is battery:)
6 0
3 years ago
Read 2 more answers
Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0
3 years ago
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