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konstantin123 [22]

3 years ago

13

When a resistor is connected to a 12-V source, it draws a 185-mA current. The same resistor connected to a 90-V source draws a 1

.25-A current. (a) Is the resistor ohmic? Justify your answer mathematically. (b) What is the rate of Joule heating in this resistor in both cases?

2 answers:

Svetach [21]3 years ago

8 0

Answer:

The resistor is not Ohmic

Explanation:

In the first case, from ohm's law: V=IR

R=V/I= 12/0.185=64.9 ohm

In the second case:

R=V/I= 90/1.25= 72ohm

The relationship between voltage and current is linear according to ohms law but that is not the case here.

Joule heating = I^2R

For case1: (0.185)^2 × 64.9 = 2.2W

For case2: (1.25)^2 ×72 =112.5W

topjm [15]3 years ago

7 0

Answer:

Explanation:

Ohm's law is define by the formula

V = IR where I is current in A, R is resistance in ohms

to calculate the resistance we use the formula above

I = 185 mA / 1000 = 0.185 A

case 1, R = V / I = 12 V / 0.185 A = 64.86 ohms

case 2, R = V / I = 90 / 1.25 A = 72 ohms

a) the resistor is not ohmic since there is no linear relationship between R and V ( as voltage increases, resistance ought to also increase)

b) rate of Joule heating in the case 1, P, power = IV = 0.185 A × 12 = 2.22 W

case 2, P = IV = 1.25 A × 90 V = 112.5 W

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a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

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a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

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$W_{f}$ - Dissipated work due to friction, measured in joules.

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$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

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$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

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$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

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