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Inessa [10]
3 years ago
11

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
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