Respirators should be worn when there is hazardous air pollution in the workplace. true or false help please

Soon Yi loves to bake, and she is making flaky pastry. Soon Yi starts with a layer of dough 222 millimeters (\text{mm})(mm)left

frozen [14]

Answer:

3 times

Explanation:

When the dough is folded, it increases by a constant factor. We can model the growth of the thickness using the exponential growth model

$T(n)=T_0(1+r)^n$

Where:

Initial thickness, $T_0$ = 2mm

Growth factor, r =8%=0.08

We want to find the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm.

i.e When $T(n)\geq 2.5$ mm$

$2(1+0.08)^n\geq 2.5\\2(1.08)^n\geq 2.5\\$Divide both sides by 2$\\\dfrac{2(1.08)^n}{2}\geq \dfrac{2.5}{2}\\\\1.08^n\geq 1.25\\\\$Change to logarithm form\\n \geq \log_{1.08}1.25\\\\n\geq \dfrac{\log 1.25}{\log 1.08} \\\\n\geq 2.9$

Therefore, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm thick is 3.

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Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca

antoniya [11.8K]

The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

$T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t}$ (1)

Where:

$T_{s}$ - Final temperature, in degrees Celsius.
$T_{o}$ - Initial temperature, in degrees Celsius.
$t$ - Time, in minutes.
$k$ - Cooling rate, in $\frac{1}{min}$ .
$T(t)$ - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when $T(t) = T_{s}$ , that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that $T_{o} = 80\,^{\circ}C$ , $t = 15\,min$ , $T_{s} = 50\,^{\circ}C$ and $T(15) = 60\,^{\circ}C$ , then the cooling rate of the substance is: