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Arturiano [62]
3 years ago
7

Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i

s y=2.0×1011n/m2. how far (δl) would such a string stretch under a tension of 1500 n? express your answer in millimeters using two significant figures.
Physics
1 answer:
laiz [17]3 years ago
4 0
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer:  15 mm
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Two speakers separated by a distance of 4.40 m emit sound. The speakers have opposite phase. A person listens from a location 3.
Hoochie [10]

Answer:

f = 147.21 Hz

Explanation:

In order to have a destructive interference, as the source emit in opposite phases, the path difference between the distance to the person, measured in a straight line from the speakers, must be equal to an integer number of wavelengths.

We need to know the distance from the listener to the other speaker, located 4.4 m from the one which is directly in front of him, which we can find using Pythagorean theorem, as follows:

l₂ = √(3)²+(4.4)² = 5.33 m

The difference in path will be, then:

d = l₂-l₁ = 5.33 m - 3.00 m = 2.33 m

For the lowest frequency that gives destructive interference, the wavelength will be highest possible, which happens when the distance is just one wavelength.

⇒ d = λ = 2.33 m

In any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

v = λ*f Κ  ⇒ f = v/λ

Taking the speed of sound as 343 m/s, and solving for f, we get:

f= 343 m/s / 2.33 m = 147.21 Hz

3 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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Sergeu [11.5K]
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W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
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Gala2k [10]

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Explanation:

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