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Tcecarenko [31]
3 years ago
10

You are trying to tune the fifth string on your guitar, which is supposed to sound the A2 note (i.e., a note two octaves below A

4). If the part of your guitar string in between where it is held down is 1.2 m and it has a mass of 0.006 kg, how much tension should you put the string under to get the right tuning?
Physics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

The value is  T =  288 \ N

Explanation:

From the question we are told that

    The length of the guitar string considered is  L = 1.2 \ m

     The mass is m = 0.006 \ kg

Generally the tension is mathematically represented as

       T = 4f ^2 m* L

Here f is the frequency of the A_2 note which is  f= 110 \ Hz

So

     T = 4* 110 ^2 * 0.006* 1.2

=>  T =  288 \ N

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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

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3 years ago
An airplane is at rest on a runway. It accelerates at 10 m/s2 for 15 seconds. How fast is it now traveling?​
KATRIN_1 [288]

Answer:

150

Explanation:

v = at

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v = 150 m/s

6 0
4 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the t
aliina [53]

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

Answer:

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

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Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

  = 45.08 N

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