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kramer
2 years ago
13

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to

the car's battery, which has an emf E = 10.0 V and an internal resistance rint = 0.04 Ω . The heating element that is immersed in the cup of water is a resistive coil with resistance R. David wants to experiment with the device, so he connects an ammeter into the circuit and measures 12.0 A when the device is connected to the car's battery. From this, he calculates the time to boil a cup of water using the device. If the energy required is 100 kJ , how long does it take to boil a cup of water?
Physics
1 answer:
jeyben [28]2 years ago
4 0

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

    V = i (r + R)

We replace we calculate

     r + R = V / i

     R = v / i - r

     R = 10/12 -0.04

     R = 0.793 Ω

We calculate the power supplied

     P = V i = I² R

     P = 12² 0.793

     P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

     P = E / t

     t = E / P

     t = 100 10³/114

     t = 877 s

The time to boil the water is 877 s

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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

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Decreased by a factor of 4.5

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F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

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8 0
2 years ago
Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.
Nady [450]

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

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Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

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the potential energy if the bow when pulled back is

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E = 1/2 * 300 * 0.5²

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The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

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v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

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v = √1500

v = 38.73 m/s

8 0
2 years ago
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