A Rocket is launched and reaches a height of 72m before falling back to Earth a) What was it's take off velocity? b) What was it
1 answer:
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Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A
Answer:
Explanation:
Recall the formula for acceleration:
, where
is final velocity,
is initial velocity, and
is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:
Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!