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bezimeni [28]
3 years ago
5

Which change will cause an increase in the electric current produced through electromagnetic induction?using more wire loops in

the solenoidusing the opposite pole of the magnetusing a weaker magnetmoving the solenoid slower?
Physics
2 answers:
miss Akunina [59]3 years ago
8 0

Answer:

using more wire loops in the solenoid

Explanation:

Here we know that rate by Faraday's law of electromagnetism

EMF = -\frac{d\phi}{dt}

Now for solenoid we know that

EMF = - L\frac{di}{dt}

now this EMF of induces need to be increased so here we need to increase

1). By increasing the self inductance(L) of the solenoid coil

2). By increasing the rate of change in flux in the coil by moving it fast

Now here inductance is increased by increasing number of turns in the coil

So here correct answer should be related to the more number of loops of the coil which will increase the electric current in the solenoid

allsm [11]3 years ago
4 0
<span>using more wire loops in the solenoid</span>
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During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima
vova2212 [387]

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

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

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

=>    F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03

=>    F  =  0.1396 \ N

8 0
3 years ago
Why does data need to be reliable
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My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
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