You are cooking beef stew. When you take the beef stew off the stove, it has a temperature of 2008F. The room temperature is 758F and the cooling rate of the beef stew is r 5 0.054. How long (in minutes) will it take to cool the beef stew to a serving temperature of 1008F

Therefore it would take about 30 minutes to cool the beef stew to a serving temperature of 1008F

Explanation:

Given that:

The initial temperature of the stew(I) is 2008° F

The room temperature(R) is 758°F

The cooling rate of the beef stew(r) = 0.054

you need the beef stew to cool to a serving temperature(F) of 1008°F

To calculate How long (in minutes) will it take to cool the beef stew to a serving temperature of 1008°F we use the equation:

Substituting values:

taking natural log of both side

Therefore it would take about 30 minutes to cool the beef stew to a serving temperature of 1008F

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards