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AnnyKZ [126]
3 years ago
14

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of

12 A through the body at 25 V for a very short time, usually about 3.0 ms. (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred
Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

(a) 300 W

(b)0.90 J

Explanation:

(a) The electric power for any circuit element is given as

P=IV

Here I is current passing in the circuit and V is the voltage.

Given I = 12 A and V =25 V.

Substitute the given values, we get

P=12 A \times25 V

P = 300 W.

Thus, the power defibrillator deliver to the body is 300 W.

(b) To calculate the electric energy use the relation between electric energy, power and time as

E=Pt

Given t = 3 ms.

Substitute the values, we get

E=300 W\times 3\times 10^-^3s

E = 0.90 J.

Thus, the energy transferred is 0.90 J.

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3 years ago
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A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
Crank

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

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The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of
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Answer:

a. 3.95\times10^{26}W

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Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum
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Answer:

d =4.77\times 10^{-5}\ m

Explanation:

given

wavelength of light  λ = 479 nm

                                   = 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°                

using                                              

condition for diffraction minimum ,

         d sinθ = m λ                        

for first minimum m = 1

       d sinθ = λ                      

therefore ,

slit width                                  

d =\dfrac{\lambda}{sin\theta}          

d =\dfrac{479\times 10^{-9}}{sin 0.575^0}

d =\dfrac{479\times 10^{-9}}{0.01}              

d =4.77\times 10^{-5}\ m                                                    

hence, the width of the slit is equal to d =4.77\times 10^{-5}\ m

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Ft-Fg= ma solve for a
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