Which of the following demonstrates a typical project benchmark progression?

If you are driving down to the sleep downgrade and you have reached the speed of 40 mph , you would apply the setrvice break unt

bezimeni [28]

Answer:

35 miles

Explanation:

When you are driving a truck that has an air brake system you have to keep in mind that when driving down a steep downgrade, the truck will automatically accelerate due to the inclination of the road, so in order to keep the speed to a controllable situation, you need to activate the service brake until you've reached the 35 miles per hour mark.

4 0

3 years ago

Explain how the objects in a battery work with its components

kotykmax [81]

Answer: Batteries have three parts

Explanation:

Batteries have three parts, an anode , a cathode , and the electrolyte. The cathode and anode the positive and negative sides at either end of a traditional battery are hooked up to an electrical circuit. The chemical reactions in the battery causes a build up of electrons at the anode.

3 0

3 years ago

If copper (which has a melting point of 1085°C) homogeneously nucleates at 849°C, calculate the critical radius given values of

____ [38]

Answer:

The critical radius is -1.30 nm

Explanation:

Temperature for homogenous nucleation of copper, $T_{H} = 849^{0} C = 849 + 273 = 1122 K$

Melting point of copper, $T_{cu} = 1085^{0} C = 1085 + 273 = 1358 K$

Latent heat of fusion, $H_{f} = -1.77 * 10^{9} J/m^{3}$

Surface free energy, $\gamma = 0.200 J/m^{2}$

Critical radius, r = ?

The formula for the critical radius is given by:

$r = \frac{2 \gamma T_{cu} }{H_{f}(T_{cu} - T_{H}) }$

$r = \frac{2 * 0.2*1358 }{(-1.77 * 10^{9}) (1358 - 1122) }$

$r = \frac{543.2 }{(-1.77 * 10^{9}) 236}\\r = -1.30 * 10^{-9} m\\r = -1.30 nm$

The critical radius is -1.30 nm

8 0

4 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

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar 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

so

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

so

$A^{*}$ 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

When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more

irakobra [83]

Answer:

how many people were asked though

Explanation:

8 0

3 years ago