Answer:
the answer is The pneumatic mechanical device can only be used as a de-icing device.
Explanation:
An ice protection system prevents the formation of ice, or enables the aircraft to shed the ice before it can grow to a dangerous thickness. Ice protection systems are designed to keep atmospheric ice from accumulating on aircraft surfaces such as wings, propellers and engine intakes.
The pneumatic mechanical device is the Pneumatic deicing boots which was invented by the Goodrich Corporation in 1923. The pneumatic boot is usually made of layers of rubber, with one or more air chambers between the layers.
Any design which utilizes either a mechanical means of breaking the bond of ice to the surface, or which operates on a periodic cycle, is necessarily a de-ice system.
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.
To calculate acceleration, divide your weight by the mass, thus the accleration is:
Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
![d=v_x t](https://tex.z-dn.net/?f=d%3Dv_x%20t)
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
![v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D%3D%5Cfrac%7B3.0%20m%7D%7B1.3%20s%7D%3D2.3%20m%2Fs)
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
![y(t) = h + v_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20%2B%20v_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
![0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s](https://tex.z-dn.net/?f=0%3Dh%2Bv_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5Cv_y%20%3D%20%5Cfrac%7B0.5gt%5E2-h%7D%7Bt%7D%3D%5Cfrac%7B0.5%289.8%20m%2Fs%5E2%29%281.3%20s%29%5E2-4.0%20m%7D%7B1.3%20s%7D%3D3.3%20m%2Fs)
So now we can find the magnitude of the initial velocity:
![v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%3D%5Csqrt%7B%282.3%20m%2Fs%29%5E2%2B%283.3%20m%2Fs%29%5E2%7D%3D4.0%20m%2Fs)