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.
<span>Voltage overcomes the resistance of the electromagnet winding to force a current through that resistance. The field strength is proportional to the coil current. More voltage pushes more current. More voltage builds up the current faster, as well as forcing it to a higher final value. </span>
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
Drawing the triangle:
H / x = tan 52.2 = 1.29
H / (4.6 - x) = tan 28.8 = .550
H = 1.29 x
H = .55 * 4.6 - .55 x
1.84 x = 2.53 combining equations
x = 1.38
4.6 - 1.38 = 3.22
Total base of triangle = 1.38 + 3.22 = 4.6
H / x = tan 52,2 = 1.29
H = 1.29 * 1.38 = 1.78 height of triangle
Check:
1.78 / 3.22 = tan 28.9
This agrees with the given value of 28.8
