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USPshnik [31]
3 years ago
9

3m/s for 12 seconds how dar would he walk

Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0
I think 36m/12s because 3×12 =36
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A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of th
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Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km

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A plate is a rigid segment Of The earth's lithosphere that moves horizontally and adjoins other plates along then zone of seismic activity.plates may include portion of both continents and ocean basin.

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an always be used to calculate the electric field. relates the electric field at points on a closed surface to the net charge en
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Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

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