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Jobisdone [24]
3 years ago
8

A viola string with a fundamental frequency of D4 (293 Hz) is generally tuned using a tension of 49.0 N. However, just before a

concert, the string breaks and only a G3 string (196 Hz) is available. The strings are the same length and are usually tuned with the same tension. What does the tension need to be on the replacement string to bring it up to the D4 frequency
Physics
1 answer:
Alborosie3 years ago
8 0

Answer:

Explanation:

For fundamental frequency in a vibrating string , the formula is

n = 1 / 2L  x  √ ( T /m₁ )

n is frequency , L is length , T is tension and m₁ is mass per unit length .

For first string ,

293 =  1 / 2L  x  √ ( 49 N  /m₁ )

For second string , let mass per unit length be m₂ .

196 =  1 / 2L  x  √ ( 49 N  /m₂ ) ------ ( 1 )

To bring its frequency back to previous one let tension be T

293  =  1 / 2L  x  √ ( T  /m₂ ) ------- ( 2 )

Dividing

293 / 196 = √ ( T  /49  )

1.4948 = √ ( T  /49  )

2.2344 = T  /49

T = 109.48 N .

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Likurg_2 [28]
The answers to the problem are as follows:

  <span>a) use f=ma. Mass of student is 836/9.8=85.3kg, force is 935-836=99N. Put it into the formula you will get a=99/85.3 

b)Use same formula. Force is 782-836=-54 so put it into formula a=-54/85.3 

c)Stopping would take longer as the acceleration is smaller 


I hope my answer has come to your help. God bless and have a nice day ahead!</span>
5 0
4 years ago
Read 2 more answers
Calculate the magnitude of the linear momentum for each of the following cases
wel

Answer:

(a)p = 1.002x10^{-20}Kg.m/s

(b)p = 0.598Kg.m/s

(c)p = 94.4Kg.m/s

(d)p = 1.77x10^{29}Kg.m/s

Explanation:

The linear momentum is defined as:

p = mv  (1)

Where m is the mass and v is the velocity

<em>a.) </em><em>A proton with mass 1.67 x10^{-27} kg moving with a velocity of 6 x 10^{6} m/s. </em>

Replacing those values in equation (1) it is gotten:

p = (1.67x10^{-27}Kg)(6x10^{6}m/s)

p = 1.002x10^{-20}Kg.m/s

So, it has a linear momentum of 1.002x10^{-20}Kg.m/s

<em>b.)</em><em> A 1.6 g bullet moving with a speed of 374m/s to the right.</em>

Notice that in this case it is necessary to express the mass of the bullet in terms of kilograms:

1.6g . \frac{1Kg}{1000g} ⇒ 1.6x10^{-3}Kg

m = 1.6x10^{-3}Kg

p = (1.6x10^{-3}Kg)(374m/s)

p = 0.598Kg.m/s

<em>c.) </em><em>A 8 kg sprinter running with a velocity of 11.8 m/s. </em>

p = (8Kg)(11.8m/s)

p = 94.4Kg.m/s

<em>d.) </em><em>Earth (m=5.98x10^{24} kg) moving with an orbital speed equal to 29700 m/s.</em>

p = (5.98x10^{24}Kg)(29700m/s)

p = 1.77x10^{29}Kg.m/s

8 0
3 years ago
On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?
Alenkasestr [34]

Answer:

<h2>2.35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

<h3>2.35 N</h3>

Hope this helps you

8 0
3 years ago
Which of the following is NOT a product of science?
Marina86 [1]

Answer:

D, None of these.

Explanation:

They have science in them but its not a product.

3 0
2 years ago
Read 2 more answers
In the United States, radioactive waste is divided into three main categories based on their activity, their heat generation pot
frozen [14]

Answer:

C.Compacting the waste and burying it in a shallow landfill.

Explanation:

It is buried in a shallow landfill because it is less reactive.

7 0
3 years ago
Read 2 more answers
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