Answer:
The answer is explained below.
Explanation:
The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.
Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.
Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation from fourth level to ground level directly.
The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.
I. Positive acceleration increases velocity. Negative acceleration decreases velocity. runner A sped up until the finish line and then slowed to a stop.
ii. Zero a acceleration implies a constant, unchanging velocity not a zero velocity. runner B achieved some velocity prior to 8s and is moving and must slow down to reach a stop.
iii. None. No aspects of this reasoning are correct. Everything she says is wrong. See iv for what/why.
iv. The sign on acceleration denotes the direction of *change in velocity* not change in direction. The sign on velocity can denote change in direction but only “forward” or “reverse” along a particular path. Cardinal direction is not indicated, generally, by the sign on velocity. It may correspond to North/South situationally but it is not an built-in feature of velocity and its sign. For example, if you are traveling with positive velocity and turn left to continue your journey you still have a positive velocity in the new direction. In fact, if you turn left again, traveling in the opposite direction as the one you started with your velocity would still be positive… in the new direction. The velocity relative to original direction could be said to be negative but that would be a confusing way to describe a journey. Maybe if you stopped the vehicle and moved in reverse, you could meaningfully say velocity was negative.
<span>Hey there!
Awesome question=)
Siobhan can place it on a regular scale(0 gravity area), or she can use the "balance scale"
</span>
I hope this helps;)
Answer:
See below
Explanation:
vf = vo + at subtract vo from both sides
vf - vo = at now divide both sides by t
(vf-vo) / t = a
Answer:
Explanation:
Step one:
given data
initial velocity u= 40m/s
time taken t=3seconds
final velocity v=?
Step two:
applying the first equation of motion
v=u-gt--- (the -ve sign implies that the arrow is against gravity)
assume g=9.81m/s^2
v=40-9.81*3
v=40-29.43
v=10.57m/s
Step three:
how high the target is located
applying
s=ut-1/2gt^2
s=40*3-1/2(9.81)*3^2
s=120-88.29/2
s=120-44.145
s=75.86m
