1. Using Strong Permanent. 2. increasing the current. 3. Decreasing the space between Magnets
Explanation:
Brainiest
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 :)
The Celsius degree is the same size as the Kelvin.
The correct choice is 'C'.
Answer:
114.86%
Explanation:
In both cases, there is a vertical force equal to the sprinter's weight:
Fy = mg
When running in a circle, there is an additional centripetal force:
Fx = mv²/r
The net force is found with Pythagorean theorem:
F² = Fx² + Fy²
F² = (mv²/r)² + (mg)²
F² = m² ((v²/r)² + g²)
F = m √((v²/r)² + g²)
Compared to just the vertical force:
F / Fy
m √((v²/r)² + g²) / mg
√((v²/r)² + g²) / g
Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:
√((12²/26)² + 9.8²) / 9.8
1.1486
The force is about 114.86% greater (round as needed).
