Answer:
<em>Well, I think the best answer will be is </em><em>1.59 g/mL Good Luck!</em>
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
(1) The harmonic number for the mode of oscillation is 3.
(2) The pitch (frequency) of the sound is 579.55 Hz
(3) The level of the water inside the vertical pipe is 0.1 m.
<h3>The harmonic number</h3>
The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.
<h3>Frequency of the wave</h3>
The pitch (frequency) of the sound is calculated from third harmonic formula;
f = 3v/4L
where;
- v is speed of sound
- L is length of the pipe
f = (3 x 340) / (4 x 0.44)
f = 579.55 Hz
<h3>level of the water</h3>
wave equation for first harmonic of a closed pipe is given as
f = v/(4L)
251.1 = 340/(4L)
4L = 340/251.1
4L = 1.35
L = 1.35/4
L = 0.34 m
level of water = 0.44 m - 0.34 m = 0.1 m
Thus, the level of the water inside the vertical pipe is 0.1 m.
Learn more about harmonics of closed pipes here: brainly.com/question/27248821
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Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.