A 0.60 kg soccer ball at rest is kicked by a player who has a mass of 50.0 kg. The final speed of the soccer ball after being ki
2 answers:
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Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M = 1.99 × 10³⁰ kg
Mass of the neutron star
M = 2( M
)
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R
² = mR
ω
²
ω² = GM
= / R
³
ω = √(GM
= / R
³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Answer:
h₍₁₎ = 495,1 meters
h₍₂₎ = 480,4 m
h₍₃₎ = 455,9 m
...
..
Explanation:
The exercise is "free fall". t =
Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s
The heights (h) according to his time (t) are found according to the formula:
h(t) = 500 - 1/2 * g * t²
Remplacing "t" with the desired time.
Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):
Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:
Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:
Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!