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3 years ago

What voltage must be applied to an 6 nF capacitor to store 0.14 mC of charge? Give answer in terms of kV.

Physics

1 answer:

levacccp [35]3 years ago

8 0

Answer:

$V=23.3kV$

Explanation:

Definition of the capacitance C, where a voltage V is applied and a charge Q is stored:

$Q=C*V$

We solve to find V:

$V=Q/C=0.14*10^{-3}C/6*10^{-9}F)=2.33*10^{4}V=23.3kV$

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Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Cu^{2+}/Cu]}=0.34V$

$E^o_{[Ag^{+}/Ag]}=0.80V$

$E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=0.80V-(0.34V)=0.46V$

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}$

$E_{cell}=0.434V$

Therefore, the cell potential for this cell 0.434 V

8 0

3 years ago

John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

sdas [7]

Average speed = distance traveled / time

average speed = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min

6 0

3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago

Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?

Bond [772]

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

7 0

3 years ago

In our Solar System, the inner planets are rocky because Choose one: A. warm temperatures in the inner disk caused the inner pla

xeze [42]

The inner planets are rocky because The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

What are rocky planets?

Rocky planets are the planet in which constituents are mostly silicate rocks or metal. They are also regarded as a planet with a solid surface.
The formation of rocky planets is said to have occurred billions of years ago and its process of formation is termed accretion. Through accretion are its constituents formed as the more it goes bigger, the higher the rising temperature and pressure in its core and the elements which have to undergo accreted heat up, melt, and spread. Through this process, heavier elements go deeper into the core of the planet and lighter elements float toward the surface.
In the formation of rocky planets, the inner portions of the disk are said to be warm from the protostar thereby resulting in the production of the heavy elements that stay there.

Examples of rocky planets are Earth or Mars

Hence, from the above, we can say that,

The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

Here,

Option A is correct.

Learn more about rocky planets here:

<u>brainly.com/question/22392798</u>

#SPJ4

3 0

2 years ago