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3 years ago

The astrometric technique of planet detection works best for

Physics

1 answer:

denpristay [2]3 years ago

6 0

The astrometric technique of planet detection works best for massive planets around nearby stars.

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Which wave has highest energy?

liubo4ka [24]

Answer:

2) A low amplitude, low-wavelength wave

8 0

3 years ago

Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.

Julli [10]

Answer:

c)The gases have the same average kinetic energy.

Explanation:

As we know that the kinetic energy of gas is given as

$K = \frac{1}{2}mv^2$

here we know that

$v = \sqrt{\frac{3RT}{M}}$

so we have

$K = \frac{1}{2}m (\frac{3RT}{M})$

now we have

$K = \frac{3}{2}n RT$

now mean kinetic energy per molecule is given as

$K_{avg} = \frac{3}{2}KT$

so this is independent of the mass of the gas

so average kinetic energy will remain same for both the gas molecules

6 0

3 years ago

Two point charges of -7uC and 4uC are a distance of 20

aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy $\mathrm{EPE}$ .

$\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}$ ,

where

The coulomb's constant $k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}$ ,
$q_1$ and $q_2$ are the sizes of the two charges, and
$r$ is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

$q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C$ ;
$q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C$ ;

Initial separation: $\rm 20\; cm = 0.20\; cm$ ;
Final separation: $\rm 90\; cm = 0.90\; cm$ .

Apply Coulomb's law:

Initial potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .