I am not so sure about this it is too difficult
Answer:
After 4 s of passing through the intersection, the train travels with 57.6 m/s
Solution:
As per the question:
Suppose the distance to the south of the crossing watching the east bound train be x = 70 m
Also, the east bound travels as a function of time and can be given as:
y(t) = 60t
Now,
To calculate the speed, z(t) of the train as it passes through the intersection:
Since, the road cross at right angles, thus by Pythagoras theorem:


Now, differentiate the above eqn w.r.t 't':


For t = 4 s:

The two-second rule.
It is a common guideline to follow while driving.
It means that any given driver should be AT LEAST two seconds behind any vehicle that is driving in front of his vehicle. It might apply for any kind of vehicle.
Answer:
71 % of the earth's surface is covered in water
Answer:

Explanation:
For the simple pendulum problem we need to remember that:
,
where
is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
,
where
is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
,
where T is the oscillation period. Now, we can easily solve for L:
