Why is symmetry important in an aircraft?

The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi

Leto [7]

Answer:

114.26

Explanation:

a)Formula for per unit impedance for change of base is

Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

kV2: New bas votlage

kVA1: given bas power

kVA2: new base power

In the question

Zpu2=??

Zpu1= 0.3

kV2=24kV

kV1= 13.8 kV

kVA2= 1MVA ×1000= 1000 kVA

kVA1=500kVA

Zpu2= 0.3(13.8/24)²×(1000/500)

Zpu2= 0.198

b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

Zbase= kV²/MVA

Zbase= 13.8²/(500/1000)

Zbase=380.88

Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

Zpu=Zactual/Zbase

0.3= Zactual/380.88

Zactual= 114.26 ohms

8 0

4 years ago

Distance in scientific notation

n200080 [17]

Answer:

(See explanation for further information)

Explanation:

a) The distances of each planet with respect to the Sun is:

Mercury - $57.909 \times 10^{6}\,mi$

Venus - $67.240\times 10^{6}\,mi$

Earth - $92.960\times 10^{6}\,mi$

Mars - $141.600\times 10^{6}\,mi$

Jupiter - $483.800\times 10^{6}\,mi$

Saturn - $888.200\times 10^{6}\,mi$

Uranus - $1,787\times 10^{6}\,mi$

Neptune - $2,795\times 10^{6}\,mi$

b) The solutions are presented below:

A. The distance between Venus and Jupiter is:

$\Delta s = 483.800\times 10^{6}\,mi-67.240\times 10^{6}\,mi$

$\Delta s = 416.560\times 10^{6}\,mi$

B. The combined distance from the Sun is:

$\Delta s = 57.909\times 10^{6}\,mi + 67.240\times 10^{6}\,mi +92.960\times 10^{6}\,mi$

$\Delta s = 218.109\times 10^{6}\,mi$

Which is less than the distance from the Sun to Neptune ( $2,795\times 10^{6}\,mi$ ).

C. The new distance of Earth is $929.60 \times 10^{6}\,mi$ ( $929,600,000\,mi$ ). Saturn would be the closest planet to Earth, whose distance:

Scientific notation:

$\Delta s = 929.600\times 10^{6}\,mi-888.200\,\times 10^{6}\,mi$

$\Delta s = 41.4\times 10^{6}\,mi$

Standard notation:

$\Delta s = 929,600,000\,mi - 888,200,000\,mi$

$\Delta s = 41,400.000\,mi$

3 0

4 years ago

4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass

Art [367]

Answer:

For mass m 1 newton 2nd law

F=m 1 a 1

5=m 1 ×10

m 1 = 2 1 kg

For mass m 2

F=m 2 a 2

5=m 2 ×20

m 2 = 4 1 kg

if tied together

Total mass =m 1 +m 2 = 1/2 +1/4=3/4kg

Now

F=M T Q T

a T = 5/m T = 5×4/3 = 3 /20m/s^2

Explanation:

8 0

3 years ago

A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m

kvasek [131]

Explanation:

Given that,

The mass of rock, m = 2.35-kg

It was released from rest at a height of 21.4 m.

(a) The kinetic energy is given by : $E_k=\dfrac{1}{2}mv^2$

As the rock was at rest initially, it means, its kinetic energy is equal to 0.

(b) The gravitational potential energy is given by : $E_p=mgh$

It can be calculated as :

$E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J$

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,

M = 0 J + 492.84 J

M = 492.84 J

Hence, this is the required solution.

6 0

3 years ago

You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?

Setler79 [48]

Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

distance = speed x time

$t_1 = \dfrac{d}{s}$

$t_1 = \dfrac{4}{30}$

t₁= 0.1333 hr

time from another 4 km trip

$t_2 = \dfrac{d}{s}$

$t_2= \dfrac{4}{50}$

t₂ = 0.08 hr

average speed

$s= \dfrac{d}{t}$

$s= \dfrac{8}{0.1333 + 0.08}$

$s= \dfrac{8}{0.2133}$

s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

6 0

3 years ago