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Aliun [14]
3 years ago
6

The ____ of a sound wave is a measure of its loudness.

Physics
1 answer:
Alisiya [41]3 years ago
8 0
The answer to this question is amplitude
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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
pentagon [3]

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

5 0
2 years ago
Which type of wave does the illustration depict?
NeX [460]

Answer is B. Longitudinal Wave

5 0
3 years ago
Read 2 more answers
Please help me with my quiz?
alina1380 [7]
The solutions would be :
1. C
2. B
3. A
4. C
5. D.
3 0
3 years ago
Jasmine is diving off a 3-meter springboard. her height in meters above the water when she is x meters horizontally from the end
olganol [36]

Answer:

5.25 m

Explanation:

Given;

The height equation h;

h=-x^2+3x+3

Where;

h = the height above water

x = horizontal distance from the end of the board

The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

2x = 3

x = 3/2

Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

h max = 21/4 = 5.25 m

8 0
3 years ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
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