Answer:
E = 440816.32 N/C
Explanation:
Given data:
Three point charge of charge equal to +3.0 micro coulomb
fourth point charge = - 3.0 micro coulomb
side of square = 0.50 m
N.m^2/c^2
Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value
So we have



[
[
]
plugging all value



E = 440816.32 N/C
<span>A boat would
produce the highest concentration of carbon monoxide in the exhaust system.
</span>Carbon monoxide<span> (CO) is a colorless, odorless, and tasteless gas that is
slightly less dense than air. It is toxic to </span>hemoglobic<span> <span>animals (both </span></span>invertebrate<span> <span>and
vertebrate, including humans) when encountered in concentrations above about 35 </span></span>ppm<span>.</span>
Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.

x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land