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Arturiano [62]
2 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]2 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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Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B
gregori [183]

Answer:

The force P required  is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

tan \ \alpha = \dfrac{CD}{BD}

where;

BD = AB-AD

Then;

tan \ \alpha = \dfrac{CD}{AB-AD}

\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )

replacing their respective values, where;

CD = 2 sin 30° m,  AB = 2m and AD = 2 cos 30° m

\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )

\alpha = tan^{-1} \bigg(3.73\bigg )

\alpha \simeq 75^0

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

T_{BC} = P

Using the force equilibrium expression along the horizontal direction.

\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0

replacing the value of \alpha \simeq 75^0

-T_{AC} \  cos 30^0 + P cos 75^0  = 0

P \ cos \ 75^0 = T_{AC} \ cos \ 30^0

P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - -  (1)

Along the vertical direction, the force equilibrium equation can be expressed as:

\sum F_y =0

-W + P \ sin \alpha + T_{AC} \ sin \ 30^0  = 0

W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0

replacing \alpha \simeq 75^0 and P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

Also, replacing W for (200 × 9.81) N

200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0

1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)

1962= T_{AC} \ (0.8660\times 3.732 + 0.5)

1962= T_{AC} \ (3.231912 + 0.5)

1962= T_{AC} \ (3.731912)

T_{AC}  = \dfrac{1962}{ \ (3.731912)}

T_{AC}  = 525.736 \ N

From P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \times0.866}{0.2588}

P = 1759.22 N

Thus, the force P required  is 1759.22 N

6 0
1 year ago
To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu
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Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

<u>B. τ = 16 Nm</u>

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The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the
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Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

P_{abs}=P_{gauge}+P_{atm} (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

P_{gauge}=\rho gh (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh

P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)

P_{atm}=95950Pa

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

S.G=\frac{\rho_{fluid}}{\rho_{water}}

\rho_{liquid}=S.G*\rho_{water}

\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}

so:

P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa

P_{abs}=137642.5 Pa

3 0
2 years ago
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