If an object is moving with____

is it true that the composition of the atmosphere changes every few kilometers as you move away from earth

laiz [17]

Yes,it's true ok? So how have you been doing

5 0

3 years ago

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

givi [52]

Answer:

M = 590.7 * 10³⁶ kg

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

3 0

3 years ago

Determine

djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

$\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F$

<h3>C1 and C2 are parallel to C3</h3>

$C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F$

<h3>C(123) is series to C5 and C6</h3>

$\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F$