Answer:
Enamel is used to coat the wires, it is the thinnest possible insulator. The coils are made up of large number of turns and enamel makes it possible to cram a lot of wires (coils) in much smaller space.
We know that the change in momentum is equals to the product of force and time that is impulse ( 
). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,
Here, u is initial velocity which is zero.
.
Thus, impulse
From Newton`s second law,
Therefore, impulse
Given, 
and 
Substituting these values, we get
Change in momentum = impulse
.
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
In this question, you're determining the time (t) taken for an object to fall from a distance (d).
The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)
d = 8,848m and g = 9.8m/s
Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)
The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)
Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)
Now for the last step, find the square root of the remaining number:
t = 42.5s
So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.
I hope this helps :)
A. The English system uses one unit for each category of measurement.
