4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:

Answer:
42 m/s
Explanation:
by definition of the velocity(speed) of a wave,
The velocity of light, v, is the product of its wavelength, λ , and its frequency, f.
V= fλ
frequency - number of occurances in a unit time
(check the graph)

Answer:
Explanation:
According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.
i.e. 

where R=resistance


whee L and d are length and Diameter
thus 
where k=constant of Variation
Answer:
charge, q = ± 1.1 mC
Given:
Capacitance, 
Voltage, V = 110 V
Solution:
The charge on the capacitor plates can be calculated by using the definition of capacitance as :
q ∝ V
where
q = charge
V = potential difference or Voltage
Therefore,
q = CV
Now, charge, q :
q = 
Therefore, the charge on the positive plate is:
q = + 1.1 mC
the charge on the negative plate is:
q = - 1.1 mC