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Zigmanuir [339]
3 years ago
8

14. A light bulb has a resistance of 8 ohms and a maximum current of 10 A How much voltage can

Physics
1 answer:
ipn [44]3 years ago
8 0

Answer:

80 volts.

Explanation:

From the question,

Applying Ohm's law

V = IR...................... Equation 1

Where V = Voltage, I = maximum current of the bulb, R = Resistance of the bulb.

Given: I = 10 A, R = ohms

Substitute these values into equation 1

V = 10×8

V = 80 volt.

Hence the voltage that can be applied is 80 volt.

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A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at rrr = 4.0 mm is πr
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Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

    The wavelength of the spherical wave is  \lambda =  2.0 \ mm =  \frac{2}{1000} = 0.002 \ m

    The first radius  is  r_1  = 4.0 \ mm  = \frac{4}{1000}  = 0.004 \ m

     The phase at that instant is  \phi _1 = \pi \ rad

     The second radius is  r_2  = 3.5 \ mm  = \frac{3.5}{1000}  = 0.0035 \ m

Generally the phase difference is mathematically represented as

          \Delta  \phi =  \phi _2 -  \phi _1

this can also be expressed as

         \Delta \phi =  \frac{2 \pi }{\lambda } (r_2 - r_1 )

So we have that

   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

substituting values

     \phi _2 -  \pi =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )

    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

   \phi _2  = 1.57 \  rad

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