A car traveled 80 kilometers in 4 hours. Its average speed was _________.
2 answers:
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Answer:
a) Final position is x = 0.90 m
b) Final position is x = 0.133 m
Explanation:
The workdone between two points is usually approximated as the area under the force-distance curve between those two points.
From the graph,
As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,
The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J
The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m
Work done under this bar = 0.4 × 0.25 = 0.1 J
Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J
So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N
0.21 = 0.18 + 0.2 (x - 0.75)
0.03 = 0.2x - 0.15
0.2x = 0.18
x = 0.9 m
Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.
b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.
From the starting point where the initial position is 0.40 m, the force here is 0.80 N
The workdone under this bar to the left is
The workdone = 0.8 (0.25 - 0.4) = - 0.12 J
Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N
- 0.19 = -0.12 + 0.6 (x - 0.25)
-0.07 = 0.6x - 0.15
0.6x = 0.08
x = (0.08/0.6) = 0.133 m
Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.
Hope this Helps!!!
As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.
Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.
As per Planck's quantum theory -
Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.
The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.
We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-
[ is the wavelength of the wave ]
It is given that wavelength =21 cm=0.21 m
The frequency is calculated as -
Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz
Here Hertz[ Hz] is the unit of frequency.