Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Explanation:
Gravitational Force Fg = GMm/r2----1
Where G is gravitational constant
M Mass of the planet, m mass of the orbit and r is the distance between the masses.
Since the circular orbit move around the planet, it means they do not touch each other.
The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.
Total distance r= (R1 + d + R2)^2---2
Recall, density rho =
Mass M/Volume V
Hence, mass of planet = rho × V
But volume of a sphere is 4/3πr3
Therefore,
Mass M of planet = rho × 4/3πr3
=4/3πr3rho in kg
From equation 1 and 2
Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Answer:
15.625 watts
Explanation:
Recall that power is defined as the worked performed per unit of time:
Power = Work / time
The work done is Force * distance, so in our case the work is:
Work = 25 M * 5 m = 125 J
Then the power will be:
Power = 125 J / 8 sec = 15.625 watts
Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
Using this 'suvat' equation
we can determine the initial velocity
Hence the initial velocity is 13.0 meters per seconds
