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Answer :
We are given that the component of the car's weight acting parallel to the slope (the grade resistance) is 112 lb when the car is on an uphill grade of [tex]$2.7^\circ$[/tex]. This component is given by
[tex]$$
\text{grade resistance} = W \sin \theta,
$$[/tex]
where [tex]$W$[/tex] is the weight of the car and [tex]$\theta$[/tex] is the angle of the grade.
Step 1. Write down the formula relating the grade resistance and the weight:
[tex]$$
W \sin \theta = 112 \text{ lb}.
$$[/tex]
Step 2. Substitute the angle [tex]$\theta = 2.7^\circ$[/tex] into the formula:
[tex]$$
W \sin (2.7^\circ) = 112.
$$[/tex]
Step 3. Solve for [tex]$W$[/tex] by dividing both sides of the equation by [tex]$\sin(2.7^\circ)$[/tex]:
[tex]$$
W = \frac{112}{\sin(2.7^\circ)}.
$$[/tex]
Step 4. Compute the sine of [tex]$2.7^\circ$[/tex]. When computed (ensuring that the angle is in degrees), we obtain
[tex]$$
\sin(2.7^\circ) \approx 0.04712389.
$$[/tex]
Step 5. Substitute the value of [tex]$\sin(2.7^\circ)$[/tex] into the equation for [tex]$W$[/tex]:
[tex]$$
W \approx \frac{112}{0.04712389} \approx 2377.59 \text{ lb}.
$$[/tex]
Thus, the weight of the car is approximately [tex]$2377.59$[/tex] lb.
[tex]$$
\text{grade resistance} = W \sin \theta,
$$[/tex]
where [tex]$W$[/tex] is the weight of the car and [tex]$\theta$[/tex] is the angle of the grade.
Step 1. Write down the formula relating the grade resistance and the weight:
[tex]$$
W \sin \theta = 112 \text{ lb}.
$$[/tex]
Step 2. Substitute the angle [tex]$\theta = 2.7^\circ$[/tex] into the formula:
[tex]$$
W \sin (2.7^\circ) = 112.
$$[/tex]
Step 3. Solve for [tex]$W$[/tex] by dividing both sides of the equation by [tex]$\sin(2.7^\circ)$[/tex]:
[tex]$$
W = \frac{112}{\sin(2.7^\circ)}.
$$[/tex]
Step 4. Compute the sine of [tex]$2.7^\circ$[/tex]. When computed (ensuring that the angle is in degrees), we obtain
[tex]$$
\sin(2.7^\circ) \approx 0.04712389.
$$[/tex]
Step 5. Substitute the value of [tex]$\sin(2.7^\circ)$[/tex] into the equation for [tex]$W$[/tex]:
[tex]$$
W \approx \frac{112}{0.04712389} \approx 2377.59 \text{ lb}.
$$[/tex]
Thus, the weight of the car is approximately [tex]$2377.59$[/tex] lb.
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