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3241004551 [841]

3 years ago

14

Why do the star constellations seem to move across the sky

2 answers:

cricket20 [7]3 years ago

7 0

Answer:

because the world moves around and around rotating in a circle and circling around the sun so as we move the stars stay the same but to our view it looks like their moving but in reality we are.

Explanation:

barxatty [35]3 years ago

5 0

Answer: These apparent star tracks are in fact not due to the stars moving, but to the rotational motion of the Earth. As the Earth rotates with an axis that is pointed in the direction of the North Star, stars appear to move from east to west in the sky.

Explanation: why do the star constellations seem to move across the sky

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A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago

olga nikolaevna [1]

Answer:

6.8 m/ s^2

Explanation:

a = f / m

a = (67N - 6N) / 9 kg

a= 6.8 m/ s^2

8 0

3 years ago

A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20

Marysya12 [62]

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

5 0

3 years ago

The equation below is used to calculate the mechanical advantage of an ideal wheel and axle.

Jet001 [13]

Answer:

b

Explanation:

6 0

2 years ago

A wheelchair ramp needs to be 8 inches high how long does it have to be to have a mechanical advantage of 5

Nastasia [14]

MA = (Effort Distance)/(Effort Resistance) = L/H

L = MA * H = 5 * 8" = 40"

8 0

3 years ago