Answer:
6.0 m/s vertical and 9.0 m/s horizontal
Explanation:
For the vertical component, we use the formula:
- Sin(34°) = <em>y</em> / 10.8
Then we <u>solve for </u><u><em>y</em></u>:
- 0.559 = <em>y</em> / 10.8
And for the horizontal component, we use the formula:
- Cos(34°) = <em>x</em> / 10.8
Then we <u>solve for </u><u><em>x</em></u><u>:</u>
- 0.829 = <em>x</em> / 10.8
So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".
Thomas Edison is the answer im 100% sure of it.
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.
Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:

where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:

where in this problem:
is the initial pressure of the gas
is the initial absolute temperature of the gas
is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:

Explanation:
It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.