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kotykmax [81]
3 years ago
5

The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a

ssociated ? whenever current flows through a conductor. Select one:
a. relay / responses
b. service / effects
c. utilization / effects
d. utilization / responses
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

utilization / effects

Explanation:

Utilization equipment are those equipment that makes use of electric energy for the purpose of chemical, electronic, lighting, heating, electro-mechanical or other alike purposes. Hence utilization best suits the first question mark in the question. Secondly, there are associated effects when current flows through a conductor, not responses.

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What would happen if you held the South Pole of one magnet near the North Pole of another magnet of the same size?
KiRa [710]

The correct choice would be

(B) The magnets would be attracted to each other.

we know that like poles repel each other and unlike poles attract each other. hence when south pole of one magnet is placed near the north pole of other magnet, the two magnets attract each other due to interaction of magnetic fields of the two magnets. the magnetic field lines originating from north pole of one magnet ends at the south of other magnet

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4 years ago
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Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is
svlad2 [7]

(a) 5.45 \cdot 10^{14} Hz

The relationship between frequency and wavelength of an electromagnetic wave is given by

c=f \lambda

where

c=3.00 \cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we are considering light with wavelength of

\lambda=5.50 \cdot 10^{-7} m

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz

(b) 1.83\cdot 10^{-15} s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

And using f=5.45 \cdot 10^{14} Hz as we found in the previous part, we can find the period of this wave:

T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s

5 0
3 years ago
The buildup of plaque on the walls of an artery may decrease its diameter from 1.1 cm to 0.90 cm. The speed of blood flow was 17
natka813 [3]

Answer

Initial radius of the artery is (1.1 cm) / 2 = 0.55 cm

final radius of the artery is (0.90 cm) / 2 = 0.45 cm

initial velocity of the blood is 17 cm/s

Using  equation of continuity is

                                      A₁v₁=A₂v₂

                                 π r₁² x v₁ = π r₂² x v₂

                                  r₁² x v₁ = r₂² x v₂

                                  0.55² x 17 =0.45² x v₂

                                        v₂=25.39 cm/s

Bernoulli's equation is

P_1 - P_2 = \dfrac{1}{2}\rho (v_2^2-v_1^2)

rho is the density of blood = 1060 kg/m^3

P_1 - P_2 = \dfrac{1}{2}\times 1060 \times (0.254^2-0.17^2)

P_1 - P_2 =18.87\ Pa

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