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larisa
4 months ago
9

A microwave oven operating at 120 volts is used to heat a sandwich if the oven draws 12.5 amperes of current for 45 seconds what

is the power dissipated by the oven?
Physics
1 answer:
kicyunya [3.2K]4 months ago
3 0
To determine the power dissipated by the oven, we can use the known voltage of 120 volts and a current of 12.5 amperes.
The power formula is P=V*I, where "V" indicates voltage and "I" represents current.
Thus, P=120*12.5
As a result, P=1500 watts.
This means the answer is 1500 watts.
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A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
Ostrovityanka [3204]
The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

     The distance the electron covers is  x = 2.50 \ m

Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

      F = 1.52 *10^{-16} \ N

Generally, the work-energy theorem is mathematically framed as

          W = \Delta KE

Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

        W = F* x *cos( \theta )

Where  \theta = 0 ^o assuming that the electron's movement aligns with the x-axis  

        So

             \Delta KE = F * x cos (0)

Inserting values

         \Delta KE = 1.52 *10^{-16} * 2.50 cos (0)

          \Delta KE = 3.8*10^{-16} J

According to the conservation of energy

       \Delta PE = - \Delta KE

Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

7 0
3 months ago
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