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3 years ago

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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

traveling a distance d.
a. What is the magnitude of the electric field?
b. What is the direction of the electric field?

1. in the direction of the electron's motion
2. opposite to the direction of the electron's motion
3. perpendicular to the direction of the electron's motion

1 answer:

KIM [24]3 years ago

7 0

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = $\frac{1}{2}mv_f^2$

But the final velocity is 0 (at rest). Hence:

Final KE = $\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0$

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

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