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3 years ago

What is a gas giant?

Physics

1 answer:

julia-pushkina [17]3 years ago

4 0

Answer:

A gas giant is a huge planet made of gases primarily hydrogen and helium. These gas giant planets include Jupiter, Saturn, Uranus, and Neptune.

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

Kazeer [188]

Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

and direction $\theta_R$ such that

$\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ$

5 0

3 years ago

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$

Hence the focal length of the convex lens is 20cm

7 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: