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sleet_krkn [62]
4 years ago
7

Which best explains why no current is induced? The wire needs to be coiled less tightly. The wire needs to be straight, not coil

ed. The magnet needs to be moved through the coils of wire. The magnet needs to be held above the coils of wire.
Physics
2 answers:
Leto [7]4 years ago
8 0

Answer : The magnet needs to be moved through the coils of wire.

Explanation : When the moves magnet towards the coil or coil towards the magnet then the change the flux and induced emf and induced current in the coil.

So, we can say that the magnet needs to be moved through the coils of wire that's why no current is induced in the coil.

Ann [662]4 years ago
7 0
The magnet needs to be held above the coils of wires

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galben [10]
The proper time would be the time measured in the rest frame of the event, so in this case the observer on mars would measure the proper time.
8 0
3 years ago
Suppose that the separation between two speakers A and B is 4.80 m and the speakers are vibrating in-phase. They are playing ide
Hatshy [7]

Answer:

X=8.44m

Explanation:

From the question we are told that

Distance b/w A&B x=4.80m

Frequency f=134Hz

Sound speed v=343m/s

Generally the equation for wavelength is mathematically given as

\lambda=v/f

\lambda/2=1/2*v/f

\lambda/2=1/2*\frac{343}{135}

\lambda/2=1.27037037

Generally the destructive interference X is mathematically given by

\sqrt{4.8^2 +X^2} -X=1.27037037\\

23.04+BC^2=X^2+1.613+2.54*X

Therefore the destructive interference is

X=8.44m

4 0
3 years ago
A wind powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of win
Hunter-Best [27]
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7 0
4 years ago
Which of the following materials is an insulator against electric current?
inna [77]
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6 0
3 years ago
Read 2 more answers
In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6
stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

                                      r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

                                      r₂= 7.5 cm

head upto water level is to rise = 60 + 5

                                          = 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q  

 π r₁² v₁ = π r₂² v₂  = 0.015

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{7.5^2}{5^2} v_2

 v_1= 2.25 v_2

 v_2 = \dfrac{0.015}{\pi r_2^2}

 v_2 = \dfrac{0.015}{\pi 0.075^2}

    v₂ = 0.848 m/s

    v₁ = 1.908 m/s

Applying Bernoulli's equation

 P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h

 P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65

 P_p= 635539.32 Pa

 P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0
4 years ago
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