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3 years ago

A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

Engineering

1 answer:

ozzi3 years ago

8 0

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by $\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]$

where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

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Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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4 0

2 years ago

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar.

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0

4 years ago

Shortly after the introduction of a new coin, newspapers published articles claiming the coin is biased. The stories were based

garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, $s = \sqrt {variance}=\sqrt {50}= 7.071068$

Z value for 108, $z =\frac {108-100}{7.071068}= 1.131371$

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

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3 years ago

If a 9V battery produces a current of 3 A through a load, what is the resistance of the load

Elden [556K]

3 ohms hope this helps :D ❤

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3 years ago