The <em><u>gravity on the Earth</u></em> has acceleration is smallest in magnitude
Given:
- Acceleration from 0 to 30 m/s in 2 s
- Deceleration from 30 m/s to 10 m/s in 1.5 s
- Gravity on Earth
To find:
The accelerations with the smallest magnitude
Solution:
- Acceleration from 0 to 30 m/s in 2 s
The acceleration in 2 seconds:

The magnitude of the acceleration is 15.
- Deceleration from 30 m/s to 10 m/s in 1.5 s
The deceleration in 1.5 seconds:

The magnitude of the deceleration is 13.4.
The value of the acceleration due to gravity = 
The magnitude of the acceleration is 9.8.
9.8 < 13.4 < 15
The <em><u>gravity on the Earth</u></em> has acceleration is smallest in magnitude.
Learn more about the acceleration here:
brainly.com/question/2437624?referrer=searchResults
brainly.com/question/7488209?referrer=searchResults
Answer:
A λ = 97.23 nm
, B) λ = 486.2 nm
, C) λ = 53326 nm
Explanation:
With that problem let's use the Bohr model equation for the hydrogen atom
= -k e² /2a₀ 1/n²
For a transition between two states we have
-
= -k e² /2a₀ (1/
² - 1 / n₀²)
Now this energy is given by the Planck equation
E = h f
And the speed of light is
c = λ f
Let's replace
h c / λ = - k e² /2a₀ (1 /
² - 1 / no₀²)
1 / λ = - k e² /2a₀ hc (1 /
² -1 / n₀²)
Where the constants are the Rydberg constant
= 1.097 10⁷ m⁻¹
1 / λ =
(1 / n₀² - 1 / nf²)
Now we can substitute the given values
Part A
Initial state n₀ = 1 to the final state
= 4
1 / λ = 1.097 10⁷ (1/1 - 1/4²)
1 / λ = 1.0284 10⁷ m⁻¹
λ = 9.723 10⁻⁸ m
We reduce to nm
λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)
λ = 97.23 nm
Part B
Initial state n₀ = 2 final state
= 4
1 / λ = 1.097 10⁷ (1/2² - 1/4²)
1 / λ = 0.2056 10⁻⁷ m
λ = 486.2 nm
Part C
Initial state n₀ = 3
1 / λ = 1,097 10⁷ (1/3² - 1/4²)
1 / λ = 5.3326 10⁵ m⁻¹
λ = 5.3326 10-5 m
λ = 53326 nm
The answer is <span>higher than.
</span><span>A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency higher than that actually being produced by the object.
Why?
</span>As the source of the waves is moving toward the observer, each of the successive wave crest<span> is emitted from a position closer to the observer than the previous wave.
Thus each wave takes slightly less time to reach the observer than the previous wave. So, the time between the arrival of successive wave crests at the observer is reduced, increasing the frequency. </span>
I have done this assignment before so I can tell you the more length that hangs over the table, the less vibration there is due to lack of tension.
Answer:
The refraction of light at the surface of water makes ponds and swimming pools appear shallower than they really are.
Explanation:
so its just the refraction of light at the surface