Where do I buy a 1997 MK4 Toyota Supra twin turbo manual for cheap

An unknown relative passes away and bequeaths upon you a small tract of land in Amherst. You decide to build a two-story storage

Ratling [72]

Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?

Explanation:

7 0

3 years ago

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

3 years ago

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

$\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}$

using the energy equation for entry and exit value :

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}$

where

$\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}$

$= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\$

$= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\$

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\$

$\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})$

L.H.S = R.H.S

7 0

3 years ago

The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter

timurjin [86]

Answer:

It’s called Expansion Joints

6 0

3 years ago

Which of the following does the electrical technician typically use?

kkurt [141]

Answer:

What are the choices?

8 0

3 years ago