Answer:
Decreases the time period of revolution
Explanation:
The time period of Cygnus X-1 orbiting a massive star is 5.6 days.
The orbital velocity of a planet is given by the formula,
v = √[GM/(R + h)]
In the case of rotational motion, v = (R +h)ω
ω = √[GM/(R + h)] /(R +h)
Where 'ω' is the angular velocity of the planet
The time period of rotational motion is,
T = 2π/ω
By substitution,
<em>T = 2π(R +h)√[(R + h)/GM] </em>
Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.
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
The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D.
The answer would then be B. constant, periodically reversing.
Answer: Rough endoplasmic reticulum, cytoplasm, inside mitochondria
Explanation: