Ask question

Ask question

All categories

3 years ago

10

A standing wave is created on a string of length 1.2 m. If the speed of the wave on the string is 60.0 m/s, what is the fundamen

tal frequency?

2 answers:

coldgirl [10]3 years ago

6 0

Answer:

Fundamental frequency in the string will be 25 Hz

Explanation:

We have given length of the string L = 1.2 m

Speed of the wave on the string v = 60 m/sec

We have to find the fundamental frequency

Fundamental frequency in the string is equal to $f=\frac{v}{2L}$ , here v is velocity on the string and L is the length of the string

So frequency will be equal to $f=\frac{v}{2L}=\frac{60}{2\times 1.2}=25Hz$

So fundamental frequency will be 25 Hz

Shalnov [3]3 years ago

3 0

Answer:

25 Hz

Explanation:

Length, L = 1.2 m

velocity, v = 60 m/s

The frequency of the fundamental is given by

f = v / 2L

f = 60 / ( 2 x 1.2)

f = 25 Hz

Thus, the fundamental frequency is 25 Hz.

You might be interested in

A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

V = 0.85

Hence

$0.85 = 1 - e^{-\frac{2.4}{\tau } }$

$- {\frac{2.4}{\tau } } = ln0.15$

$\tau = 1.265 s$

7 0

3 years ago

A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below

siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

$W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J$

Power is given by

$P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds$

The time taken by the horse to pull the professor is 125.83672 seconds

6 0

3 years ago

The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the

Advocard [28]

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

5 0

3 years ago

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

3 years ago

Suppose an air bubble is trapped in the eudiometer before starting the experiment. After the experiment is finished, the resulti

seraphim [82]

Answer:

Explanation:

The resulting valie would be too large

8 0

2 years ago