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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

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2. A student drives 7.8-km trip to school and averages a speed of
Alekssandra [29.7K]

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

\displaystyle V=\frac{x}{t}

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec

The total time is:

t=239.26\ sec+211.96\ sec=451.22\ sec

\boxed{t=451.22\ sec}

The average speed is:

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\boxed{\displaystyle V=34.57\ m/s}

6 0
3 years ago
A curve of radius 166 m is banked at an angle of 11°. An 736-kg car negotiates the curve at 81 km/h without skidding. Neglect th
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Answer:

F_n = 7509.33\ N

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given,

radius of curve = 166 m

angle of the banked road = 11°

mass of car = 736 Kg

speed of the curve = 81 km/h

                                = 81 x 0.278 = 22.52 m/s

normal force acting on the tires

on tire there will be two force acting on it

first one will be force acting due to weight and the other force acting on the tire is due to centripetal force.

F_n = m g cos \theta+ \dfrac{mv^2}{r} sin \theta

F_n = 736 \times 9.8 \times cos 11^0+\dfrac{736 \times 22.52^2}{166} sin 11^0

F_n = 7080.28+429.047

F_n = 7509.33\ N

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