Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
The correct answer for the question that is being presented above is this one: "(a)4." Suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly in phase and maximum <span>reinforcement occurs, in 1 second, there will be 4 beats.</span>
Answer:
A. 
B. t = 50 s
Explanation:
A. The vectorial equation of the person who is getting closer to the other person is:
r: position vector
v: speed vector = 6m/s i (if you consider the motion as a horizontal motion)
Then, you replace and obtain:
B. The time is:
d: distance to the observer = 300m
v: speed of the person on the car = 6.00 m/s
The answer would be a radio wave
