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3 years ago

Helpp me pleassee....

Physics

1 answer:

lilavasa [31]3 years ago

4 0

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

$f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz$

Now fundamental wavelength is :

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m$

So, the fundamental wavelength of the vibrating string is 1.7 m.

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Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge

Katen [24]

Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

7 0

3 years ago

At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 226 N/C. At a distance r2 fro

Svetradugi [14.3K]

Answer:

r2/r1 = 1.3

Explanation:

Electric field is given as:

E = kq/r²

At a distance r1,

226 = kq/(r1)² - - - - - - - - - - - - - (1)

At a distance r2,

134 = kq/(r2)² - - - - - - - - - - - - - (2)

From (1),

kq = 226 * (r1)²

From (2),

kq = 134 * (r2)²

Equating and then solving,

134 * (r2)² = 226 * (r1)²

(r2)²/(r1)² = 226/134

(r2)²/(r1)² = 1.687

=> r2/r1 = 1.3

6 0

3 years ago

Read 2 more answers

A mustang, has an average velocity of 33 m/s while covering a course that is 21 miles long. (1 mile = 1609 m). How long did it t

saw5 [17]

Answer:

t = 1023.9 seconds

Explanation:

Given that,

The average velocity of a Mustang, v = 33m/s

Distance, d = 21 miles = 33789 m

Let the driver takes t seconds. So,

Speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{33789}{33}\\\\t=1023.9\ s$

So, it will take 1023.9 seconds to complete the course.

3 0

2 years ago

A ray of light contains two colors: red with wavelength of 660 nm and blue with wavelength 470 nm. The ray passes through two na

densk [106]

Answer:

The distance on the screen between the first-order bright fringes for each wavelength is 3.17 mm.

Explanation:

Given that,

Wavelength of red = 660 nm

Wavelength of blue = 470 nm

Separated d= 0.30 mm

Distance between screen and slits D= 5.0 m

We need to calculate the distance for red wavelength

Using formula for distance

$y=\dfrac{\lambda D}{d}$

Where, D = distance between screen and slits

d = separation of slits

Put the value into the formula

$y=\dfrac{660\times10^{-9}\times5.0}{0.30\times10^{-3}}$

$y=11\ mm$

For blue wavelength,

Put the value into the formula again

$y'=\dfrac{470\times10^{-9}\times5.0}{0.30\times10^{-3}}$

$y'=7.83\ mm$

We need to calculate the distance on the screen between the first-order bright fringes for each wavelength

Using formula for distance

$\Delta y=y-y'$

$\Delta y=11-7.83$

$\Delta y=3.17\ mm$

Hence, The distance on the screen between the first-order bright fringes for each wavelength is 3.17 mm.

4 0

4 years ago

• A cable is suspended over a pulley. A 5.0 kg mass is attached to

Galina-37 [17]

Answer:

Tension= $\frac{20g}{3}$ (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

$10g-T = 10a$

$T-5g=5a$

Now adding them we get,

$5g=15a$

$a=\frac{g}{3}$

Substituting them back in the equation we get,

$10g-10(\frac{g}{3} ) = T$

$T= \frac{20g}{3}$

3 0

3 years ago