Answer:
- Wind resistance made decrease in speed
-Gravity/Mass made decrease in velocity
Explanation:
The velocity and acceleration of the particle moving round the circle is;
<em><u>Velocity = 162.12 m/s</u></em>
<em><u>Velocity = 162.12 m/sAcceleration = 6.873 × 10^(-5) m/s²</u></em>
We are given;
Radius of circle; 382400 km = 382400000 m
Time; t = 27.3 days = 27.3 × 86400 s = 2358720 s
Now, formula for velocity is;
Velocity = distance/time
Thus;
I) velocity = 382400000/2358720
Velocity = 162.12 m/s
II) Acceleration is centripetal acceleration and is given by the formula;
a = v²/r
a = 162.12²/382400000
a = 6.873 × 10^(-5) m/s²
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Answer:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:
where
is the gravitational constant
is the mass of the telescope
is the mass of the Earth
is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:
Going to college and passing all your classes
Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
