1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
adoni [48]
3 years ago
7

On a cello, the string with the largest linear density (1.58 10-2 kg/m) is the C string. This string produces a fundamental freq

uency of 65.4 Hz and has a length of 0.805 m between the two fixed ends. Find the tension in the string.
Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

F=175.2N

Explanation:

The relationship of between fundamental frequency, density of the string, length and tension of the string is:

L=\frac{1}{2f} \sqrt{\frac{F}{\frac{L}{m} } }

where L/m is the linear density.

make F the subject of the formula

F=4L^{2}f^{2}  \frac{L}{m}

substitute the values

F=4*0.805^{2} *65.4^{2} *(1.58*10^{-2} )\\F=175.2N

You might be interested in
Why are the peaks opposite in direction?
blagie [28]

Answer:

A

Explanation:

Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field

4 0
3 years ago
A student measured the volume of three rocks. The diagram above shows the
victus00 [196]

Answer:

option B is right

...............

4 0
2 years ago
Use your calculator to evaluate -3.7 meter/second-13.9 meter/second 21.4 second-72 second
cluponka [151]

Answer :

(-3.7 meter/second) - (13.9 meter/second) = -17.6 meter/second

(21.4 second) - (72 second) = -50.6 second

Explanation :

(1) As we are given the expression :

(-3.7 meter/second) - (13.9 meter/second)

Now we have to evaluate this expression, we get:

⇒ -17.6 meter/second

(2) As we are given the expression :

(21.4 second) - (72 second)

Now we have to evaluate this expression, we get:

⇒ -50.6 second

6 0
3 years ago
Which characteristic does an object with a constant acceleration always have?
Alexus [3.1K]

By definition, speed is the integral of acceleration with respect to time.

We have then:

v = \int\limits^t_0 {a} \, dt

As the acceleration is constant, then integrating we have:

v = a*t + vo

Where,

vo: constant of integration that corresponds to the initial velocity

We observe then that the speed varies linearly when the acceleration is constant .

Therefore, for constant acceleration, the velocity is changing.

Answer:

an object with a constant acceleration always have:

A. changing velocity

6 0
3 years ago
Read 2 more answers
I don't know the answer but I think that it is yes I just don't want to get it wrong
mrs_skeptik [129]
Yes your answer should be true.
8 0
3 years ago
Read 2 more answers
Other questions:
  • What did the greeks call the mineral they found and why?
    6·1 answer
  • The oxygen isotopic composition of ocean water is measured by determining the ratio of 18O to 16O, expressed as 18O/16O. The iso
    11·1 answer
  • How does Scientific theories differ from scientific laws
    8·1 answer
  • How does the science of heat transfer differ from the science of thermodynamics?
    10·1 answer
  • Pluto has a shape that is nearly round,and it orbits the sun,it has five known moons.why is it called a dwarf planet and not a p
    10·1 answer
  • What is inertia? (1 point)
    10·1 answer
  • How many minutes will it take a car to go from a stop to 33 km/hr if it accelerates at 10 km/hr²
    6·2 answers
  • Earth's gravity acts upon objects with a steady force of __________. A. 8. 9 meters per second B. 9. 8 meters per minute C. 8. 9
    12·1 answer
  • Please help with this i am not good with this
    15·1 answer
  • a new planet is found with a density one third as much at earth and a radius twice that of earth. what is the acceleration due t
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!