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anzhelika [568]
3 years ago
5

What are some examples of a solute

Physics
1 answer:
Leto [7]3 years ago
7 0
Anything that can be dissolved/mixed into a solvent. Sugar, salt, acetic acid, etc
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You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
2 years ago
Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he
11111nata11111 [884]

Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
2 years ago
A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?
Sergeu [11.5K]

Answer:

216 miles

Explanation:

you times 72 by three

7 0
3 years ago
Read 2 more answers
1.)State which definition of power below is correct.
Natalija [7]

C

kW, watts, kilowatts, W

6 0
3 years ago
Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.
anzhelika [568]
Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
E^0=E_{cat}^0-E_{an}^0
where E_{cat}^0 is the standard potential at the cathode, while E_{an}^0 is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: E_{Cu}^0=+0.34 V, while at the anode we have zinc: E_{Zn}^0=-0.76 V. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
E^0=+0.34 V-(-0.76 V)=+1.1 V

2) To calculate E^0 at any temperature T, we should use Nerst equation:
E^0(T)=E^0- \frac{R T}{z F} \ln  \frac{[Zn]}{[Cu]}
where 
R=8.31 J/(K mol)
T=473.15 K is the temperature in our problem
z=2 is the number of electrons transferred in the cell's reaction
F=9.65\cdot 10^4 C/mol is the Faraday's constant
[Zn] and [Cu] are the molar concentrations of zinc and in copper, and in our problem we have [Zn]=10[Cu].
Using all these data inside the equation, and using E^0=+1.1 V, in the end we find:
E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V
8 0
2 years ago
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