Quản trị học là gì ? ý nghĩa của quản trị học với thực tế xã hội

Find the Rectangular form of the following phasors?

almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45, 30, 60, 90, -34, -56, 20, -42, -65, -15

P=10, P=5, P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10, θ = 45° rectangular coordinates

x = r cosθ , y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

= 7.07

y =P sinθ = 10 sin 45°

= 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos 30° = 4.33

y = 5 sin 30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos 60° = 12.5

y = 25 sin 60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos 90° = 0

y = 54 sin 90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos (-34°) = 53.88

y = 65 sin (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos (-56)° = 53.12

y = 95 sin (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos 20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos (-42)° = 5.94

y = 8 sin (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos (-65)° = 14.79

y = 35 sin (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos (-15)° = 144.88

y = 150 sin (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0

2 years ago

The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama De

Leto [7]

Answer:

<:

Explanation:

8 0

2 years ago

Read 2 more answers

Give two disadvantages of a moving coil Meter

Temka [501]

1.Only suitable for dc
2.more expensive than moving iron type
3. Easily damaged

7 0

2 years ago

The future goal of health information is appropriate access to ubiquitous, personalized, specific information relevant to the si

erastovalidia [21]

I thinks it’s a A.True

3 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$