Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
Answer:
292796435 seconds ≈ 300 million seconds
Explanation:
First of all, the speed of the car is 121km/h = 33.6111 m/s
The radius of the planet is given to be 7380 km = 7380000 m
From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec
If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have
w(vehicle) = 9.78 x w(planet)
w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec
To find the period of the planet's rotation; we use the equation
w(planet) = 2π÷T
Where w(planet) is the angular velocity of the planet and T is the period
From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds
Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds
Answer:
-2 m/s^2
Explanation:
Acceleration is equal to the slope of the graph. You just find the slope of that section. The rise is -20 and the run is 10, so you get -2.
Explanation:
It is given that,
The horizontal speed of a cliff diver, 
It reaches the water below 2.00 s later, t = 2 s
Let 
is the distance where the diver hit the water. It can be calculated as follows :
Let 
is the height of the cliff. It can be calculated using second equation of motion as follows :
So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.
