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3 years ago

14

For this graph to the right, what is the velocity of the object? Show your work.

1 answer:

mr Goodwill [35]3 years ago

7 0

I dont see a mark where it is, if there was a line it would be marking the constant speed of it. There is just time(s) which probably stands for speed, and position(m) which probably stands for miles. There is a measurement of 50, down to 10. Please give me more explanation to this question and I may be able to help!

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Do you think seismographs predict earthquakes or measure earthquakes, explain your answer?

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Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinet

Natasha_Volkova [10]

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

Ions accelerated through voltage, V = 10.3 volts

The work done to change the position of singly charged gas ions is given by the relation :

W = q x V

Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = $\frac{3}{2}kT$

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

$\frac{3}{2}kT = qV$

Rearrange the above equation in terms of T :

$T= \frac{2qV}{3k}$

Substitute the suitable values in the above equation.

$T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }$

T = 7.96 x 10⁴ K

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3 years ago

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yanalaym [24]

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3 0

4 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

$v_{d}$ = drift velocity

Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

$t=6402.31\ sec$

$t=106.7\ minute$

Hence, The time is 106.7 minute.

7 0

3 years ago