The energy of an electron as it is ejected from the atom can be calculated from the product of the Planck's constant and the frequency of the light energy. We can calculate the wavelength from the frequency we can calculate. We do as follows:
E = hv
4.41 x 10-19 = 6.62607004 × 10<span>-34 (v)
v = 6.66x10^14 /s
wavelength = speed of light / frequency
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wavelength = 3x10^8 / 6.66x10^14
wavelength = 4.51x10^-7 m = 450.75 nm
(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
Answer: The first answer for the first problem, and the 2nd answer for the second problem
Explanation: For the first one, if it is absolute zero, the molecules would not move at all.
For the second one, the temperature of the sample will increase due to the movement.
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
6^2 + 8^2 = 36 + 64 = 100
sqrt(100) = 10 m/s northwest