<span>We calculate the electric field as follows:
r = </span>√<span>(3)/6 x 19 cm = .05484 m
The angle for the triangle would be 30 on each side.
tan(30) = r/(L/2)
E' = kQ/{r*sqrt[(L/2)^2 + r^2]} = (8.99e9 x 15e-9) / {.05484 * sqrt[(.19/2)^2 + .003]}
</span>E' <span>= 22413 N/C
The value above is the electric field strength for a single rod at the center.
|E'| = 22413 N/C
E = 2|E'|sin(30) + |E'| = 49000 N/C</span>
To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.
From the theory we could consider that the energy change is subject to
Where
Initial Energy
Energy loses
Replacing we have that
Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV
A: 0.7 -0.8 and 0hrs
B: 0.2 - 0.4 and 1.0 -1.1 hrs
C: 0-0.2 hrs and 0.8 - 1.0 hrs
D: 0.4 - 0.7 hrs
Answer:
48.26 m
Explanation:
time to goes up (till stop for a while in the air - maximum height)
vt = vo + a t
0 = 15 + g . t
0 = 15 + (-9.8) . t
9.8t = 15
t = 1.531 s
so the time left to goes down is
4.0 - 1.531 = 2.469 s
height from the top of building can find it by using
vo =√(2gh)
15 = √(2)(9.8).h
15² = 19.6h
h = 225/19.6 = 11.48 m
so the distance of maximum height to the ground is
t = √(2H/g)
2.469 = √(2H/9.8)
2.469² = 2H/9.8
6.096 = 2H/9.8
2H = 6.096 x 9.8 = 59.74 m
so the vertical distance of the building (or the building height's is)
H - h = 59.74 - 11.48 = 48.26 m
If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.
If the mass of the sun is 2x, at least one planet will fall into the habitable zone. if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully.
If the mass of the sun is 3x, at least one planet will fall into the habitable zone if I place a planet in orbits 672, and 7 and all planets will orbit the sun successfully.
