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3 years ago

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mina [271]3 years ago

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Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

Hannah just finished building a house of cards that stands four stories high. She is worried that it will fall down. Which of th

Tju [1.3M]

A.) <span>If no unbalanced force acts upon the house of cards, then it will remain standing forever.

[ Other Statements doesn't make any sense in Physics, they can be true in some situation & can't be false, option A is the only answer ]
Hope this helps!</span>

6 0

3 years ago

Read 2 more answers

100 joule of heat produced each 1 secomd on 4ohm find the potential difference

irakobra [83]

Answer:

20 V

Explanation:

Power is 100 J/s or 100 W.

We know that P = IV = $\frac{V^{2} }{R}$ .

Isolate the potential difference. V = $\sqrt{RP}$ = $\sqrt{100 * 4}$ = 20 V

8 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

Calculate the wavelength in centimeters of radar energy at a frequency of 10 GHz. What is the frequency in gigahertz of radar en

konstantin123 [22]

Explanation:

(a) Frequency of radar energy, $f = 10\ GHz =10^{10}\ Hz$

The relation between frequency and wavelength is given by :

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8}{10^{10}}$

$\lambda=0.03\ m$

or

$\lambda=3\ cm$

(b) If wavelength, $\lambda=25\ cm=0.25\ m$

$c=f\lambda$

$f=\dfrac{c}{\lambda}$

$f=\dfrac{3\times 10^8}{0.25}$

$f=1.2\times 10^9\ Hz$

or

f = 1.2 GHz

Hence, this is the required solution.

7 0

2 years ago