By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.
Hope this helps
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Explanation :
A power amplifier is used to amplify electric signals i.e. to increase the low power signal to higher powers.
A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base junction is forward biased.
This transistor can be used as a power amplifier because it gives a much larger output current. The gain of an amplifier shows the amount of amplification. It is the difference between the input and the output signals.
Answer:
α = 395 rad/s²
Explanation:
Main features of uniformly accelerated circular motion
A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.
There is tangential acceleration (at) and is constant.
at = α*R Formula (1)
where
α is the angular acceleration
R is the radius of the circular path
There is normal or centripetal acceleration that determines the change in direction of the velocity vector.
Data
R = 0.0600 m :blade radius
at = 23.7 m/s² : tangential acceleration of the blades
Angular acceleration of the blades (α)
We replace data in the formula (1)
at = α*R
23.7 = α*(0.06)
α = (23.7) / (0.06)
α = 395 rad/s²
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The correct option would be A. solar.
We also use solar energy to produce electricity.
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Answer:

Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, 
Let the N number of electrons have charge -40 nC, such that,
Now, mass of one electron = 
Therefore, mass of N electrons = 
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is 
Then,

It is the required fraction of mass of the droplet.