Answer:
Explanation:
fundamental frequency, f = 250 Hz
Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.
the formula for the frequency is given by
.... (1)
Now the length is doubled ans the tension is four times but the mass remains same.
let the frequency is f'
.... (2)
Divide equation (2) by equation (1)
f' = √2 x f
f' = 1.414 x 250
f' = 353.5 Hz
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
Answer:
0.83 g/cm³
Explanation:
The volume of the alcohol is ...
(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³
The density of water is about 1 g/cm³, so the volume of 9 g of it is ...
(9 g)/(1 g/cm³) = 9 cm³
The total volume is ...
42.0253 cm³ +9 cm³ = 51.0253 cm³
The total mass is ...
33.2 g + 9 g = 42.4 g
So, the resulting density is ...
(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³
The resulting mixture has a density of about 0.83 g/cm³.
_____
<em>Additional comment</em>
Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.
The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...
(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59
The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.
We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV
