Answer:
Bent knees
Explanation:
Depending the height your jumping from, if you were jumping from a high structure it's best if you jump and land on knees so you dont sprang your ankles from jumping and landing on your straight legs, you could even pull a muscle if you land on straight legs.
Answer:

Explanation:
Given that

We know that acceleration a given as




We know that



So the magnitude of force F

Answer:
16,506 ft²
Explanation:
There are different ways you can divide the area using rectangles and circles. One way is to find the area of the entire width and length, then subtract the empty areas in the corners.
If we take the empty areas and put them together, we find their area is the area of a square minus the area of a circle.
A = (2r)² − πr²
A = 4r² − πr²
A = (4 − π) r²
So the area of the rink is:
A = WL − (4 − π) r²
A = (85)(200) − (4 − π) (24)²
A ≈ 16,506 ft²
Answer:
The hill should be not less than 0.625 m high
Explanation:
This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that
is constant, being U the potential energy and K the kinetic energy


When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.
The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We can solve for h:

The hill should be not less than 0.625 m high