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erica [24]
3 years ago
8

iven a 36.0 V battery and 14.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in serie

s.
Physics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

0.367A = Current of both resistors

For resistor 1: 1.89W; For resistor 2: 11.3W

Explanation:

When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:

R = 14.0Ω + 84.0Ω = 98.0Ω

Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):

V = RI

V / R = I

36.0V / 98.0Ω = I

<h3>0.367A = Current of both resistors</h3><h3 />

Power is defined as:

P = I²*R

For resistor 1:

P = 0.367A²*14.0Ω = 1.89W

For resistor 1:

P = 0.367A²*84.0Ω = 11.3W

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Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out
PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is  

\lambda _2 = (6.48,6.52) \ cm

Explanation:

From the question we are told that

     The wavelength of the note is \lambda  =  6.50 \ cm = 0.065 \ m

     The difference in beat frequency is \Delta  f = 17.0 \ Hz

     

Generally the frequency of the note played by the guitar that is in tune is  

        f_1 = \frac{v_s}{\lambda}

Where v_s is the speed of sound with a constant value v_s  =  343 \ m/s

       f_1 = \frac{343}{0.0065}

      f_1 = 5276.9 \ Hz

The difference in beat is mathematically represented as

       \Delta  f =  |f_1 - f_2|

Where f_2 is the frequency of the sound from the out of tune guitar

     f_2 =f_1  \pm \Delta f

substituting values

      f_2 =f_1 + \Delta f

      f_2 = 5276.9 + 17.0  

     f_2 = 5293.9 \ Hz

The wavelength for this frequency is

      \lambda_2 = \frac{343 }{5293.9}

     \lambda_2 = 0.0648 \ m

    \lambda_2 = 6.48 \ cm

For the second value of the second frequency

     f_2 =  f_1 - \Delta f

     f_2 = 5276.9 -17

      f_2 = 5259.9 Hz

The wavelength for this frequency is

   \lambda _2 = \frac{343}{5259.9}

   \lambda _2 = 0.0652 \ m

   \lambda _2 = 6.52 \ cm

8 0
3 years ago
fuel was consumed at a certain rate of 0.05Kg\s in a rocket engine and ejected as a gas with a speed of4000m\s . Determine the t
ivann1987 [24]

Answer:

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Explanation:

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m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

Thrust = (0.05\ kg/s)(4000\ m/s)

<u>Thrust = 200 N</u>

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