Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
Resistivity = resistance * Area / Length
resistance = ??
rho = 2.30 * 10^3 ohms meters.
l = 23.5 cm = 0.235 meters.
diameter = 2.62 cm = 0.0262 m
Radius = diameter / 2 = 0.0131 m
Area = pi r^2 where pi = 3.14
Area = 3.14 * 0.0131^2 = 0.000539 m^2
2.30 * 10^3 = R * 0.000539 /0.235
2.30 * 10^3 * 0.235/0.000539 = R
R = 1002783 which is about 1 meg ohm, but I'll use the calculation here.
E = 1000 V
R = 1 002 783 ohms
I = ???
E = I * R
1000 = I * 1002783 We should really round that resistor to 1 megohm or 1 * 10^6 ohms.
I = 1*10^3 / 1 * 10^6
I = 1 * 10^-3 amp = 1 milliamp. Pretty reasonable all things considered.
Based on this scatter plot, I can say that in general, test scores vary directly with study time. In other words, in a population large enough to reveal the direct relationship, the more time a student spends studying, the higher that student's test scores will be.
I didn't need a scatter plot to reveal that to me. That's practically in the category of Law given to Moses at Sinai.
Answer:
Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto
Explanation:
