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2 years ago

Define prisms. How to draw true shape of a prism?

Engineering

1 answer:

aksik [14]2 years ago

4 0

Answer:

In geometry, a prism is a polyhedron comprising an n-sided polygonal base, a second base which is a translated copy of the first, and n other faces joining corresponding sides of the two bases. All cross-sections parallel to the bases are translations of the bases.

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Techm digital definition comprises of

KonstantinChe [14]

Answer:

Below is the response to the given question:

Explanation:

The relevant services supplied through TechMahindra Digital Services Provider are among the different options given in n inquiry. This is a digital company that has offered its customers an end-to-end solution that digitalizes all the requirements for client operations. It offers digital solutions, services cloud-based, digital marketing strategies, and then all client needs.

8 0

2 years ago

A car travels from A, due north to a town B 4 km away. It then travels due east until it arrives town C 5 km from B. determine t

morpeh [17]

Answer:

A to C = 6.4 km

Explanation:

A to B = 4 km

B to C = 5 km

A to C = using pythagorean theorem

a² + b² = c²

a = A to B = 4

b = B to C = 5

c = A to C

c² = 4² + 5²

c = 6.4 km (A to C)

8 0

2 years ago

Currently, the lost time of each stage is 4 seconds, and intersection critical v/c ratio is set to be 0.75 to avoid cycle failur

KIM [24]

Find the solution in the attachments

Note: Question was incomplete, so the complete question is added in the attachments.

7 0

2 years ago

Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e

Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

from the voltage divide rule

$V_{in} = V_0 * \frac{R_2}{R_2 + R_f}$

$\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20$

$\frac{R_f}{R_2} = 19$

Select the nearest answer

obtained gain = $\frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19$

Expected gain = $\frac{V_0}{V_{in}} = 20$

∴ error = | $\frac{19 - 20}{20}$ | × 100

= 1/20 × 100

= 5%

6 0

2 years ago

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys

vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

$=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

$=\frac{1}{\sqrt3 }$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

3 0

2 years ago