Answer:
The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is
ft/ sec
Explanation:
Given:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec . 5 = 125 ft
x(5) = 10 ft/sec . 5 = 50 ft
Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

Lets find the derivative of distance with respect to time

Substituting the values of h(t) and x(t) and simplifying we get,



=
=
ft / sec
Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV

ΔU =18.79 mJ
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

where
are the mass of the child, the boat and the package, respectively.
are the velocity of the package and the boat after throwing.



Anions: negatively charged atoms
Cations: positively charged atoms (CAT-ions are PAWS-itively charged)
Anode: positively charged electrode that allows electrical current flow (positive end of battery)
Cathode: negatively charged electrode that allows electrical current flow (negative end of batter)
Answer:
i. The pressure of due to the water, <em>P</em>, is given according to the following equation;
P = ρ·g·h
Where;
ρ = The density of the water (a constant) = 997 kg/m³
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the water (minimum h = h₁, maximum h = h₂)
The pressure is directly proportional to the water height, and we have;
The pressure, <em>P</em>, will be maximum when the water height, <em>h</em>, is maximum or h = h₂, which is the level DC
ii. The thrust = The force acting on the body = Pressure × Area
The maximum areas exposed to the water are on side AB and DC
However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC
Explanation: