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masya89 [10]
3 years ago
10

Service conductors installed as unjacketed multiconductor cable shall ahv a minimum clearance of ft from windows that are design

ed to be opened dorr sproches balconues laddedrs stairs fire escapes or similar locationA. TrueB. False
Physics
1 answer:
natima [27]3 years ago
8 0

Answer:

A. True

Explanation:

Service conductors are electrical conductors tapped from the service point of the utility company to the service disconnecting means at the consumer's end.

Service conductors installed as unjacketed multi-conductor cable shall have a minimum clearance of 3ft from windows that are designed to be opened, doors, balconies, ladders, porches, stairs, fire escapes, or similar locations.

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Elastic potential energy
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You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am
emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

1 - (v₀/343) = 0.9141

v₀/343 = 1 - 0.9141 = 0.0859

v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour

7 0
3 years ago
Please help (will mark brainliest)
serg [7]

Answer:

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7 0
2 years ago
Read 2 more answers
When you measure the boiling point of mercury, you are investigating a ___. a.chemical change b.chemical property c.physical cha
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D. physical property

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3 0
3 years ago
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

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3 years ago
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