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3 years ago

What is the electric potential 16.0 cm from a 4.00 μC point charge?

Physics

1 answer:

AveGali [126]3 years ago

7 0

Answer:

the potential at a distance of 16 cm from the charge $4\mu C$ will be 225000 volt

Explanation:

We have given charge $q=4\mu C=4\times 10^{-6}C$

Distance between the charge r = 16 cm = 0.16 m

Electric potential is given by $V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}$ , here K is constant which value is $9\times 10^9Nm^2/C^2$

So potential $V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt$

So the potential at a distance of 16 cm from the charge $4\mu C$ will be 225000 volt

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Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

REY [17]

Answer:

a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

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3 years ago

What happens when an object is dropped?

katrin2010 [14]

When an object is dropped, tossed, or kicked, as long as it is not laying on the ground, it accelerates downward, because of the force of gravity acting on it.

3 0

3 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$

Where:

$V$ is the speed of travel of the Moon around the Earth

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=384400(10)^{3} m$ is the distance from the center of the Earth to the center of the Moon

Solving:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}$

$V=1018.26 m/s$ This is the speed of travel of the Moon around the Earth

5 0

2 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the

saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

$R=\alpha \frac{l}{s}$

Were:

R is resistance

$\alpha$ is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area ( $s=\pi r^{2}$ ).

So if the lenght and radius are doubled, the equation goes as follows:

$R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }$

So finally because the circular area is a square function, the resulting equation is half of the one before.

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3 years ago

Consider a particle with unit charge q, and mass m, in a constant magnetic field B directed along the positive z–axis. The parti

max2010maxim [7]

Answer:

it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

Explanation:

Magnetic field is along z axis while velocity is in x-z plane

so we will have

$F = q(\vec v \times \vec B)$

so here we can say

$F = q(u\hat i + w\hat k) \times (B \hat k)$

so we will have

$F = quB(-\hat j)$

so here the net force on the charge is perpendicular to its x directional velocity along - Y direction

So due to this component of motion it will move along a circle while other component of the motion will remain uniform always

So here it is combination of two parts

1) Uniform circular motion

2) Uniform motion

So we can say that it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

4 0

3 years ago