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Calculate the energy consumed by a bread toaster given voltage, current, and time.

Physics

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3 parts
d

a) 5.78×104 J\text{a) } 5.78 \times 10^4 \text{ J}

13

a) C/3\text{a) } C/3

14

a) S\text{a) S}

Step-by-step solution

9 steps

Part 12· 4 steps

Energy consumed by a bread toaster

  1. Identify the given values.

    Current, I=4.0 AI = 4.0 \text{ A}
    Voltage, V=240.0 VV = 240.0 \text{ V}
    Time, t=1 minutet = 1 \text{ minute}

  2. Convert time to seconds.

    t=1 minute×60 seconds1 minute=60 st = 1 \text{ minute} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 60 \text{ s}
  3. Use the formula for electrical energy consumed.

    Energy (EE) is given by the product of voltage (VV), current (II), and time (tt):

    E=V×I×tE = V \times I \times t
  4. Substitute the values and calculate the energy.

    E=240.0 V×4.0 A×60 sE = 240.0 \text{ V} \times 4.0 \text{ A} \times 60 \text{ s} E=960 W×60 sE = 960 \text{ W} \times 60 \text{ s} E=57600 JE = 57600 \text{ J}

    Express in scientific notation:

    E=5.76×104 JE = 5.76 \times 10^4 \text{ J}

    Comparing with the given options:

Part a

5.78×104 J5.78 \times 10^4 \text{ J}

Part b

3.60×103 J3.60 \times 10^3 \text{ J}

Part c

3.87×104 J3.87 \times 10^4 \text{ J}

Part d

a) 5.78×104 J\text{a) } 5.78 \times 10^4 \text{ J}

6.48×105 J6.48 \times 10^5 \text{ J}

The calculated value 5.76×104 J5.76 \times 10^4 \text{ J} is closest to option (a). There might be a slight rounding difference in the option.

The energy consumed by the toaster is a) 5.78×104 J\boxed{\text{a) } 5.78 \times 10^4 \text{ J}}.


Part 13

a) C/3\text{a) } C/3

Equivalent capacitance of three capacitors in series

  1. Identify the given information.

    Number of capacitors, n=3n = 3
    Capacitance of each capacitor, C1=C2=C3=CC_1 = C_2 = C_3 = C
    Connection type: Series

  2. Recall the formula for equivalent capacitance in series.

    For capacitors connected in series, the reciprocal of the equivalent capacitance (CeqC_{eq}) is the sum of the reciprocals of individual capacitances:

    1Ceq=1C1+1C2+1C3\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
  3. Substitute the given capacitances and solve for CeqC_{eq}.

    1Ceq=1C+1C+1C\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} 1Ceq=1+1+1C\frac{1}{C_{eq}} = \frac{1+1+1}{C} 1Ceq=3C\frac{1}{C_{eq}} = \frac{3}{C}

    To find CeqC_{eq}, take the reciprocal of both sides:

    Ceq=C3C_{eq} = \frac{C}{3}

    The equivalent capacitance is a) C/3\boxed{\text{a) } C/3}.


Part 14

a) S\text{a) S}

Hooke's Law proportionality

  1. Recall Hooke's Law.

    Hooke's Law states that the force (FF) required to extend or compress a spring by some distance (xx) is directly proportional to that distance, provided the elastic limit is not exceeded.
    Mathematically, this is expressed as:

    F=kxF = kx

    Where kk is the spring constant.

  2. Apply Hooke's Law to the given problem.

    The problem states that the force needed to elongate an elastic object by an amount "S" is proportional to "S".
    According to Hooke's Law, the force (FF) is directly proportional to the elongation (SS).

    F∝SF \propto S

    The force needed to elongate an elastic object by an amount "S" is proportional to a) S\boxed{\text{a) S}}.

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