Answer:
<u>For M84:</u>
M = 590.7 * 10³⁶ kg
<u>For M87:</u>
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:
making m the subject of the formula in the equation above to calculate the mass of the black hole:
.............(1)
<u>For M84:</u>
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 = 590.7 * 10³⁶ kg
<u>For M87:</u>
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 = 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.
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The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s