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-BARSIC- [3]
4 years ago
6

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the

standard A pitch, transverse waves with a wavelength of 15.3 cm travel along the string. Calculate the tension T in the string.
Physics
1 answer:
blsea [12.9K]4 years ago
5 0

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

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A sports car has an average acceleration of 5.81 m/s2. How long does it take for the car to reach 60.0 mi/h, if it starts from r
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A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi
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Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

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