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3 years ago

Interstellar dust causes stars to appear bluer than they actually are

1 answer:

4 0

Im not sure what your question is but I will just elaborate. Interstellar dust absorbs more red light than blue light, making stars appear bluer than their true color.

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In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

6 0

3 years ago

In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.

garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

$\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...$

Find the equivalent resistance of the right branch of the circuit:

$R_{eq} = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms$

$\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411$

6 0

2 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

6 0

2 years ago

Three common elements that can reorient their electrons into magnetic domains and become magnetic are iron, nickel, and ________

ohaa [14]

Answer:

Cobalt

Explanation:

8 0

2 years ago

In a lab, the mass of object A is 2.5 kg. Object A weighs:

dangina [55]

Answer:

25N

Explanation:

Assuming the lab is on earth:

w = mg = 2.5 (9.81) = 25N

8 0

3 years ago