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2 years ago

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An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The cur

rent delivered to the coffee pot is 8.5 A when it is plugged into a 120 V electrical outlet. If the specific heat capacity of water is 4186 J/kgC° , approximately how long does it take to boil all of the water?

1 answer:

MariettaO [177]2 years ago

6 0

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$

is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

$\Delta T=100-21\ ^0C=79\ ^0C$

So,

$\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ$

Heat Supplied $Q=VIt$

where $i=current$

$V=Voltage$

$8.5\times 120\times t=2.4\times 4186\times 79$

$t=778.10 s$

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Answer:

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Explanation:

The problem can be solved by using the following SUVAT equation:

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For the diver in the problem, we have:

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By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

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Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

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Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

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Part D

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energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

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Answer:

The car strikes the tree with a final speed of 4.165 m/s

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Explanation:

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Once we have the initial speed, we can isolate the final speed from following equation:

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Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

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mart [117]

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Answer:

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$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

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The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

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So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

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