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2 years ago

THE COMPUND INTEREST ON RS 30,000AT 7% PER ANNUM IS RS 4347 THE PERIOD IN YEARS

Engineering

2 answers:

Georgia [21]2 years ago

6 0

Answer:

2 years

1st year ci 2100

2nd year ci 2100 +147(7%of 2100)

timofeeve [1]2 years ago

4 0

Answer:

what wym

Explanation:

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The time delay of a long-distance call can be determined by multiplying a small fixed constant by the number of communication li

aliina [53]

Answer:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

Remove all leaves of T. Let the remaining tree be T1.
Remove all leaves of T1. Let the remaining tree be T2.
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
If Tk has only one node, that is the center of T. The diameter of T is 2k.
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

Explanation:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

Remove all leaves of T. Let the remaining tree be T1.
Remove all leaves of T1. Let the remaining tree be T2.
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
If Tk has only one node, that is the center of T. The diameter of T is 2k.
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

4 0

2 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

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Explanation:

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