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AleksandrR [38]
3 years ago
12

How do you raise the frequency of a guitar string?

Physics
1 answer:
marusya05 [52]3 years ago
3 0
You pull the string harder
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How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
AnnyKZ [126]

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

7 0
3 years ago
Read 2 more answers
Frequency, period and wavelength<br> 11th grade high school physics
zzz [600]

Answer:0.38

Explanation:

the formula is f = c / λ

so f= 2.5/6.5

and that equals 0.38 46 and so on so i just rounded it

6 0
3 years ago
Can we write names while writing conversation in board exam​
lora16 [44]

Answer:

ya we can write the imaginary character's name .

So that we  can identify these imaginary people, as we cannot simply write the conversation and leave it .

Or maybe sometimes the reader will get confused as there is no name for the two people .

So, i suggest that you should write the names

Explanation:

You can even ask to your class teacher for further clarification

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3 years ago
Come all ruj-jxgn-qua​
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Answer:

this app is for solving doubts not sending links ok

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3 years ago
A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

                 I = mr^2

Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

      Now expressing the above statement mathematically

            potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}

               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

Substituting values

                v = \sqrt{9.81 * 1.7}

                  v=4.08m/s

4 0
3 years ago
Read 2 more answers
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