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otez555 [7]
2 years ago
9

A paper clip is made of wire 0.5 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diameter

, calculate the longitudinal engineering and true strains that the wire has undergone during processing
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
3 0

Answer:

longitudinal engineering strain = 624.16

true strain is 6.44

Explanation:

given data

diameter d1 = 0.5 mm

diameter d2 = 25 mm

to find out

longitudinal engineering and true strains

solution

we know both the volume is same

so

volume 1 = volume 2

A×L(1) = A×L(2)

( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)

( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)

0.1963 ×L(1) = 122.71 ×L(2)

L(1) / L(2) = 122.71 / 0.1963 = 625.16

and we know longitudinal engineering strain is

longitudinal engineering strain = L(1) / L(2)  - 1

longitudinal engineering strain = 625.16  - 1

longitudinal engineering strain = 624.16

and

true strain is

true strain = ln ( L(1) / L(2))

true strain = ln ( 625.16)

true strain is 6.44

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When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance?
grandymaker [24]

Answer:

<h2>1.5 ohms</h2>

Explanation:

Power is expressed as P = V²/R

R = resistance

V = supplied voltage

Given P = 600W and V = 30V

R = V²/P

R = 30²/600

R = 900/600

R = 1.5ohms

magnitude of its resistance is 1.5ohms

3 0
1 year ago
A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

\frac{di}{dt} = 2.5 A/s

so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

4 0
1 year ago
A large crate is at rest on a ramp at a loading dock
VARVARA [1.3K]

Answer:

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Explanation:

4 0
1 year ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
1 year ago
A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis
snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

\rho = \frac{RA}{L}

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length  of the wire

Rearranging, we have;

R= \frac{\rho L}{A}

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = \pi \cdot \left (\frac{D}{2}   \right )^{2} \times \frac{1}{4}  = \pi \cdot \frac{D^{2}}{16} = A/4

Therefore, the new resistance, R₂ can be expressed as follows;

R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times  \frac{\rho L}{A}

Hence, the new resistance R₂ =  2×R, that is the resistance will be doubled.

8 0
2 years ago
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