Explanation:
It is given that,
An electron is released from rest in a weak electric field of, 
Vertical distance covered, 
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

.............(1)
Electric force is
and force of gravity is
. As both forces are acting in downward direction. So, total force is:



Acceleration of the electron, 


Put the value of a in equation (1) as :


v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.
C is the right image for that biological process.
Answer:
The average velocity is 0.15 m/s
Explanation:
Use the definition of average velocity as the distance traveled divided the time it took.
Since the movement was on the plane from the origin (0, 0) to the point (-30, 20) corresponding to 30 m west and 20 m north, we calculate the distance using the distance between two points on the plane:

Then the magnitude of the average velocity can be estimated via the quotient between distance divided time, but since the units required are meters per second, we first convert the four minute time into seconds: 4 * 60 = 240 seconds.
Then the average velocity becomes:

Answer: <em>Around each new moon and full moon, the sun, Earth, and moon arrange themselves more or less along a line in space. Then the pull on the tides increases, because the gravity of the sun reinforces the moon's gravity. ... This is the spring tide: the highest (and lowest) tide. Spring tides are not named for the season.</em>
The time taken for the tiny saliva to travel is 0.55 second.
The horizontal distance traveled at speed of 4 m/s is 2.2 m.
The horizontal distance traveled at speed of 20 m/s is 11 m.
<h3>
Time of motion of the tiny saliva</h3>
The time taken for the tiny saliva to travel is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- g is the acceleration due to gravity
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
Substitute the given parameters and solve for time of motion;
t = √(2 x 1.5 / 10)
t = 0.55 second
<h3>Horizontal distance traveled at speed of 4 m/s</h3>
X = Vx(t)
X = (4 m/s)(0.55)
X = 2.2 m
<h3>Horizontal distance traveled at speed of 20 m/s</h3>
X = (20)(0.55)
X = 11 m
Learn more about time of motion here: brainly.com/question/2364404
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