Answer:
10 m/s
Explanation:
Momentum before collision = momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v
64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v
60 kg m/s = (6 kg) v
v = 10 m/s
Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R = 
Gauge Pressure inside, 
Blood Pressure outside, 
Now,
Change in pressure, 
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:


And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.
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:
sin 2θ = 1 θ=45
Explanation:
They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation
R = Vo² sin 2θ / g
Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.
We calculate the distance traveled for different angle
R = vo² Sin (2 15) /9.8
R = Vo² 0.051 m
In the table are all values in two ways
Angle (θ) distance R (x)
0 0 0
15 0.051 Vo² 0.5 Vo²/g
30 0.088 vo² 0.866 Vo²/g
45 0.102 Vo² 1 Vo²/g
60 0.088 Vo² 0.866 Vo²/g
75 0.051 vo² 0.5 Vo²/g
90 0 0
See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL
mass gram, time sec, temp kelvin, vol liter, dens grams/cm3