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Schach [20]
3 years ago
14

A quadraphonic car stereo operates on electric current provided by the car's 12V battery and is connected in series. Each channe

l of the stereo, which feeds the electrical signal to one of the stereo's four speakers, has a resistance of 4.1 oums. How much current is in the circuit of each stereo channel?
Physics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

2.93 A

Explanation:

From Ohm's law V = IR where V = voltage = 12 V, I = current = ? and R = resistance = 4.1 Ω.

So, the current in each quadraphonic stereo circuit is I = V/R = 12 V/4.1 Ω = 2.93 A

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pogonyaev
Let h =  distance (m) to the water surface.

Initial velocity, u  = 0 (because the stone was dropped).
Use the formula
 h = ut + (1/2)gt^2
  where g = 9.8 m/s^2  (acc. due to graity)
              t = time (s)

h = (1/2)*(9.8)*(3^2) = 44.1 m
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3 years ago
Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the va
Illusion [34]

Answer:

True  b and c

Explanation:

In an RLC circuit the impedance is

         Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ]     }

examine the different phrases..

a) False. The maximum impedance is the value of the resistance  

b) True. Resonance occurs when  

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               w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum  

d) False. In resonance the inductive and capacitive part cancel each other out  

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6 0
3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

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Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

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