Answer:
A fundamental theory that provides a description of the physical properties of nature at the scale of atoms and subatomic particles.
Explanation:
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
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:

Explanation:
The given data is as follows.
Angular velocity (
) = 2.23 rps
Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
= 2.23 revolutions per second
=
= 14.01 rad/s
Now, tangential speed will be calculated as follows.
Tangential speed, v =
= 0.379 x 14.01
= 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.
To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>
Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>