the answer is concave lens
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer: Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.
Explanation:
Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
![(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]](https://tex.z-dn.net/?f=%28P_%7B1%7D%20%2B%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B1%7D%20%29%3D%28P_%7B2%7D%20%2B%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B2%7D%20%29%5C%5CP_%7B1%7D%20%3DP_%7B2%7D%20%5C%5Cv_%7B1%7D%3D0%5C%5Ch_%7B2%7D%20%3D0%5C%5Cv_%7B2%7D%3D%5Csqrt%7B0.54%2A9.81%2A2%7D%5C%5Cv_%7B2%7D%3D3.25%5Bm%2Fs%5D)