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4 years ago

In a string instrument, a thicker string has an

Physics

1 answer:

Rama09 [41]4 years ago

7 0

In a string instrument, thicker strings will produce a lower sound and octave than the thinner strings.

Even if the strings are the same length, thicker strings will still produce a lower sound.

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two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

6 0

3 years ago

Please need help on this

Rasek [7]

Im pretty sure this is A. i may be incorrect, give brainliest if im correct please!

7 0

4 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.76 m. After the mass is added, the length is remeasured

ElenaW [278]

Answer:

$1242337 N/m^2$

Explanation:

The cross section area of the cable is

$A = \pi r^2 = \pi * 0.025^2 = 0.002 m^2$

Let g = 9.81m/s2. The stress acting on the cable when mass is added is

$\sigma = F/A = mg/A = 35*9.81/0.002 = 174867 Pa$

The strain when the cable is stretched from 4.76 to 5.43 m is

$\epsilon = \frac{\Delta L}{L} = \frac{5.43 - 4.76}{4.76} = 0.14$

So the young modulus of the cable is

$E = \sigma / \epsilon = 174867 / 0.14 = 1242337 Pa = 1242337 N/m^2$

4 0

3 years ago

Read 2 more answers

In general, as the energy of a sound wave increases, A. the sound gets softer and then louder. B. the sound gets louder. C. the

Vadim26 [7]

The sound gets louder, like hitting a metal pole on a thin metal beam, the energy increases as the sound does.

5 0

4 years ago

Read 2 more answers

I provided the question above.

VikaD [51]

Answer:

Explanation:

since it is connected in parallel combination

use this formula

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2}$

$\frac{1}{R} = \frac{1}{2} + \frac{1}{4}$

$\frac{1}{R} = \frac{4+2}{2}$

$\frac{1}{R} = \frac{6}{2}$

$\frac{1}{R} = 3$ ohm

therefore resistence = 3 ohm

then we should find power

P = VI

P = 12*3

P = 24 watt

now to find current use formula power = current * voltage

24 = current * 12

24/12 = current

2 = current

therefore current is 2 ampere (A).

to find potential difference (emf) use formula

V = IR

V = current * resistence

V = 2 * 3

V = 6 volt .

therefore potential difference is 6 volt.

6 0

3 years ago

In a string instrument, a thicker string has an​

In a string instrument, a thicker string has an