Answer:
F = 53153.36[N]
Explanation:
In order to solve this problem, we must first use the principle of conservation of energy which is transformed from potential energy to kinetic, in this way we can determine the velocity at which the person enters the water.
![E_{pot}=E_{kin}\\m*g*h=\frac{1}{2}*m*v^{2}](https://tex.z-dn.net/?f=E_%7Bpot%7D%3DE_%7Bkin%7D%5C%5Cm%2Ag%2Ah%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D)
where:
m = mass = 100 [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 9 [m]
v = velocity [m/s]
Now replacing we can determinate the velocity.
![v^{2}=2*g*h\\v=\sqrt{2*g*h}\\v=\sqrt{2*9.81*9}\\v = 13.28[m/s]](https://tex.z-dn.net/?f=v%5E%7B2%7D%3D2%2Ag%2Ah%5C%5Cv%3D%5Csqrt%7B2%2Ag%2Ah%7D%5C%5Cv%3D%5Csqrt%7B2%2A9.81%2A9%7D%5C%5Cv%20%3D%2013.28%5Bm%2Fs%5D)
Then we can calculate the momentum which can be calculated as the product of force by time, this momentum is also equal to the product of mass by velocity.
![P=m*v\\and\\P =F*t\\F*t=m*v](https://tex.z-dn.net/?f=P%3Dm%2Av%5C%5Cand%5C%5CP%20%3DF%2At%5C%5CF%2At%3Dm%2Av)
Now replacing:
F = impact force [N]
t = time = 0.025 [s]
m = 100 [kg]
v = velocity = 13.28 [m/s]
![F*0.025=100*13.28\\F=53153.36[N]](https://tex.z-dn.net/?f=F%2A0.025%3D100%2A13.28%5C%5CF%3D53153.36%5BN%5D)
Answer:
A capacitor
Explanation:
Because it can store electric energy when disconnected from its charging circuit. Commonly used in electronic devices to maintain power supply while batteries change.
Hope this helps! :)
Answer:
and ![a_y=0\ \text{m/s}^2](https://tex.z-dn.net/?f=a_y%3D0%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
Magnitude of accleration is
and the direction is ![0^{\circ}](https://tex.z-dn.net/?f=0%5E%7B%5Ccirc%7D)
Explanation:
![t_1=2\ \text{s}](https://tex.z-dn.net/?f=t_1%3D2%5C%20%5Ctext%7Bs%7D)
![v_x=1\ \text{m/s}](https://tex.z-dn.net/?f=v_x%3D1%5C%20%5Ctext%7Bm%2Fs%7D)
![v_y=3\ \text{m/s}](https://tex.z-dn.net/?f=v_y%3D3%5C%20%5Ctext%7Bm%2Fs%7D)
![t_2=2.5\ \text{s}](https://tex.z-dn.net/?f=t_2%3D2.5%5C%20%5Ctext%7Bs%7D)
![v_x=4\ \text{m/s}](https://tex.z-dn.net/?f=v_x%3D4%5C%20%5Ctext%7Bm%2Fs%7D)
![v_y=3\ \text{m/s}](https://tex.z-dn.net/?f=v_y%3D3%5C%20%5Ctext%7Bm%2Fs%7D)
Average acceleration in the different axes
![a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2](https://tex.z-dn.net/?f=a_x%3D%5Cdfrac%7B%5CDelta%20v_x%7D%7B%5CDelta%20t%7D%5C%5C%5CRightarrow%20a_x%3D%5Cdfrac%7B4-1%7D%7B2.5-2%7D%5C%5C%5CRightarrow%20a_x%3D6%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
![a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2](https://tex.z-dn.net/?f=a_y%3D%5Cdfrac%7B%5CDelta%20v_y%7D%7B%5CDelta%20t%7D%5C%5C%5CRightarrow%20a_y%3D%5Cdfrac%7B3-3%7D%7B2.5-2%7D%5C%5C%5CRightarrow%20a_y%3D0%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
The components of the acceleration is
and ![a_y=0\ \text{m/s}^2](https://tex.z-dn.net/?f=a_y%3D0%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
The magnitude of acceleration
![a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2](https://tex.z-dn.net/?f=a%3D%5Csqrt%7Ba_x%5E2%2Ba_y%5E2%7D%5C%5C%5CRightarrow%20a%3D%5Csqrt%7B6%5E2%2B0%5E2%7D%5C%5C%5CRightarrow%20a%3D6%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
Direction
![\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctan%5E%7B-1%7D%5Cdfrac%7Ba_y%7D%7Ba_x%7D%5C%5C%5CRightarrow%20%5Ctheta%3D%5Ctan%5E%7B-1%7D%5Cdfrac%7B0%7D%7B6%7D%5C%5C%5CRightarrow%20%5Ctheta%3D0%5E%7B%5Ccirc%7D)
The magnitude of accleration is
and the direction is
.
Answer:
Vx= 11.0865(m/s)
Vy= 6.4008(m/s)
Explanation:
Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:
V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)
The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:
Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)
And for the same theorem the speed on the Y axis will be:
Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)