Answer:
10259.6 m
Explanation:
We are given that
Radius of small wheel,r=0.17 m
Radius of large wheel,r'=0.92 m
Initial velocity,u=0
Time,t=2.7 minutes=162 s
1 min=60 s
Velocity,v=10m/s
Time,t'=13.7 minutes=822 s
Time,t''=4.1 minutes=246 s
Substitute the values
Substitute the values
Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m
let the plane is flying at some angle theta with west towards north
Now we can use the components
now since plane has to fly towards west so net speed in north must be equal to speed in south for air
So we can say
so plane has to fly in direction 14.5 degree North of west
(a)
At a distance r=R from the centre of the planet, there is no effect due to the outer shell: so, the gravitational field strength at r=R is only determined by the gravity produced by the core of the planet.
So, the strength of the gravitational field is given by
where
G is the gravitational constant
M = 6.24 × 10^24 kg is the mass of the core of the planet
R = 4.11 × 10^6 m is the radius of the core
Substituting into the equation, we find
(b)
at distance r=3R from the centre, the particle feels the effect of gravity due to both the core of the planet and the outer shell between R and 2R.
So, we have to consider the total mass that exerts the gravitational attraction at r=3R, which is the sum of the mass of the core (M) and the mass of the shell (4M):
M' = M + 4M = 5M
Therefore, the gravitational acceleration at r=3R will be
And susbstituting
g = 24.6 m/s^2
found in the previous part, we find