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maksim [4K]
4 years ago
14

A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cros

s section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Physics
1 answer:
EleoNora [17]4 years ago
5 0

Answer:

The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross sectionA = 3\times10^{-3}\ m^2

Length of copper wire = 3.5 m

Cross sectionA = 4\times10^{-5}\ m^2

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}

Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}....(I)

The young's modulus for copper wire

Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}

\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}

Hence, The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

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A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a
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Answer:

E   = (0.56 \times 10^8 ) r   \   \ N/c

Explanation:

Given that:

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To find  the electric field for r < R by using Gauss Law

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For r < R

Q_{enclosed}=(\rho) ( \pi r^2 ) l

E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}

E= \dfrac{\rho ( r)}{2 \varepsilon_o}

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\varepsilon_o = 8.85 \times 10^{-12}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E   = (0.56 \times 10^8 ) r   \   \ N/c

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A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o
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Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

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v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

6 0
3 years ago
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