Answer:
direct current hope this helped : )
<span>k = 1.7 x 10^5 kg/s^2
Player mass = 69 kg
Hooke's law states
F = kX
where
F = Force
k = spring constant
X = deflection
So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration.
F = kX
F/X = k
115 kg* 9.8 m/s^2 / 0.65 cm
= 115 kg* 9.8 m/s^2 / 0.0065 m
= 1127 kg*m/s^2 / 0.0065 m
= 173384.6154 kg/s^2
Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2
Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So
X/0.39 cm = 115 kg/0.65 cm
X = 44.85 kg/0.65
X = 69 kg
The player masses 69 kg.</span>
Force is all the same, but pressure will be higher on 3cm
