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.
Explanation:
It is given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, 
Speed of light, 
Let t is the time interval required for the strong interaction to occur. The speed is given by :
So, the time interval required for the strong interaction to occur is 
. Hence, this is the required solution.
Um this doesn't make since to me since you did not clearly state your awnser
Answer:
f₂ = 468.67 Hz
Explanation:
A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:
The higher frequency tone = f₁ = 470 Hz
No. of beats = n = 4 beats
Time period = t = 3 s
The lower frequency note = Frequency of Friend's Trombone = f₂ = ?
Beat Frequency = fb
So, the formula for beats per second or beat frequency is given as:
fb = n/t
fb = 4 beats/ 3 s
fb = 1.33 Hz
Another formula for beat frequency is:
fb = f₁ - f₂
f₂ = f₁ - fb
f₂ = 470 Hz - 1.33 Hz
<u>f₂ = 468.67 Hz</u>
