Initial velocity u = 40
Angle at launch = 55 degrees
At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
Maximum height = 54.7 m
We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s
Distance from castle to trebuchet = utcos55 = 40 x 6.68 x 0.573 = 153.1 m
Answer:
v = 31.32 [m/s]
Explanation:
To solve this problem we must use the principle of energy conservation, which tells us that potential energy is converted into kinetic energy or vice versa. The potential energy can be calculated by the product of mass by gravity by height.
where:
Epot = potential energy [J]
m = mass = 25 [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 50 [m]
Now replacing:
![E_{pot}=25*9.81*50\\E_{pot}= 12262.5[J]](https://tex.z-dn.net/?f=E_%7Bpot%7D%3D25%2A9.81%2A50%5C%5CE_%7Bpot%7D%3D%2012262.5%5BJ%5D)
When the rock falls the potential energy is converted into kinetic energy.
where:
Ek = kinetic energy [J]
v = velocity [m/s]
Now clearing v:
![v^{2} =\frac{E_{k}*2}{m}\\v=\sqrt{(2*12262)/25}\\v = 31.32 [m/s]](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%5Cfrac%7BE_%7Bk%7D%2A2%7D%7Bm%7D%5C%5Cv%3D%5Csqrt%7B%282%2A12262%29%2F25%7D%5C%5Cv%20%3D%2031.32%20%5Bm%2Fs%5D)
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Answer: I observed that the electric field strength is the same at all points between the plates. The value of the field is 70 volts per meter. This is exactly 100 times the applied voltage. The electric field lines point from the positive plate to the negative plate, as the downward arrow on the detector shows.
Explanation:
the sample answer, don't directly copy it!
