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3 years ago

An inside measurement can be taken with a tape measure by

Engineering

1 answer:

Cerrena [4.2K]3 years ago

3 0

This is about method of taking inside measurement with a tape.

<u><em>Option A is the correct answer.</em></u>

Inside measurement is usually taken with the aid of an inside micrometer. This will give us an exact measurement. This is because when making use of an inside micrometer for inside measurement, the total length of the inside micrometer is equal to the overall length of whatever is being measured.

We have seen how the inside micrometer is used whereby the entire body is also included in whatever is being measured. Applying this same concept to using a tape, we can simply say that we will add the tape length to the measuring case to get the inside measurement.

From the definitions and comparisons above, we can see that we will have to add the length of the tape measure case when taking inside measurements and other options aside Option A are not correct.

Read more at; brainly.com/question/12709703

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Viết một đoạn văn nói về tình bạn

inn [45]

Answer:

Một người bạn luôn quan trọng trong cuộc sống của chúng ta, và mọi người đều thích sự bầu bạn của một người bạn. Tình bạn thật sự rất khó để có được. Trải qua mọi khó khăn, thất bại, người bạn thủy chung sẽ luôn sát cánh. Họ sẽ quan tâm đến bạn mọi lúc, và có được một tình bạn đích thực là một món quà thực sự Có được một người bạn đồng hành trong cuộc sống của bạn là rất quan trọng. Một người có thể hiểu được cảm xúc của bạn, ủng hộ bạn và sát cánh bên bạn trong những lúc tốt và xấu ngay cả khi mọi người quay lưng lại với bạn, một người như thế thật đáng quý biết bao. Đây là người mà chúng tôi gọi là bạn thân nhất của mình.

Explanation:

I hope that helps you

7 0

3 years ago

Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential

Deffense [45]

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/( $A^{*}$ ₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃ in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403 $e^{j51.34 }$

c) value of A₃ in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937 $e^{j81.34 }$

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁ = 5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( $\frac{5}{4}$ )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A $e^{j\beta }$

so, A₂ in exponential form will be;

A₂ = 6.403 $e^{j51.34 }$

exponential form of A₂ = 6.403 $e^{j51.34 }$

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( $\frac{8 }{9.196}$ )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°

A₃ = 12.19∠41.02°

Therefore, value of A₃ in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/( $A^{*}$ ₂)

giving your answer in exponential form

we know that $A^{*}$ ₂ is the complex conjugate of A₂

$A^{*}$ ₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A $e^{j\beta }$

$A^{*}$ ₂ = 6.403 $e^{-j51.34 }$

also

A₁ = 6∠30

we convert to polar form

A₁ = 6 $e^{j30 }$

so A₅ = A₁/( $A^{*}$ ₂)

A₅ = 6 $e^{j30 }$ / 6.403 $e^{-j51.34 }$

A₅ = (6/6.403) $e^{j(30+51.34) }$

A₅ = 0.937 $e^{j81.34 }$

Therefore A₅ in exponential form is 0.937 $e^{j81.34 }$

6 0

3 years ago

How much work is required to move an electron from the positive terminal to the negative terminal of a 12 V battery? (Knight 21.

Ray Of Light [21]

Answer:

work which is required to move an electron from the positive terminal to the negative terminal is

-4μJ

W = -4μJ

Explanation:

Work required to move the charge

W = qΔV

initial point V₁ = 150V

Final point V₂ = -50V

W = q(V₂ -V₁)

= 20 × 10⁹(-50 - 150)

W = -4μJ

work which is required to move an electron from the positive terminal to the negative terminal is

-4μJ

6 0

3 years ago

The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr

Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

$(\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}$

Since d = 2r

Then:

$(\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}$

$(\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}$

$(\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}$

$(\sigma_ x) = 11917000.38$

$(\sigma_ x) = 11.917 \times 10^6 \ Pa$

$(\sigma_ x) = 11.917 \ MPa$

$\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa$