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2 years ago

6

given a circuit powered at 12V with R1, R2, R3 respectively of 10,20,30 Ohm, determine R4 in such a way that the Wheatstone brid

ge is in equilibrium and then calculate voltages and currents as for reference exercise.

1 answer:

elena55 [62]2 years ago

5 0

Answer:

The balanced condition for Wheatstones bridge is

Q

P

=

S

R

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R

1

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R

2

=R+S

=20+30=50Ω

Net resistance of the network as R

1

and R

2

are in parallel

i=

R

V

=

50

6×3

=0.36 A.

Explanation:

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An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(a). We need to calculate the induced emf

Using formula of emf

$E = Blv\sin\theta$

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(b). We need to calculate the power dissipated through the resistor

Using formula of power

$P=I^2 R$

Put the value into the formula

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

6 0

3 years ago

Read 2 more answers

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pickupchik [31]

Answer:

G

Explanation:

4 0

2 years ago

Select all the correct answers.

myrzilka [38]

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3 years ago

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a

MrRissso [65]

Answer:

$h>\dfrac{5}{2}R$

Explanation:

Given that

Height = h

Radius = R

From energy conservation

$KE_A+U_A=KE_B+U_B$

At point B

The minimum speed to complete the the circle

$V_B=\sqrt{gR}\ m/s$

So the kinetic energy at point B

$KE_B=\dfrac{1}{2}mV^2$

$KE_B=\dfrac{1}{2}mgR$

$KE_A+U_A=KE_B+U_B$

$0+mgh=\dfrac{1}{2}mgR+2mgR$

Without falling off at the top (point B)

$0+mgh>\dfrac{1}{2}mgR+2mgR$

$mg(h-2R)>\dfrac{1}{2}mgR$

$g(h-2R)>\dfrac{1}{2}gR$

$h>\dfrac{5}{2}R$

6 0

3 years ago

Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima

Svet_ta [14]

Answer:

$5.241\ \text{m/s}$

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = $9.81\ \text{m/s}^2$

h = Altura = $1.4\ \text{m}$

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}$

La velocidad máxima que puede alcanzar el coche es $5.241\ \text{m/s}$ .

8 0

3 years ago