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sergij07 [2.7K]
3 years ago
15

During an all-night cram session, a student heats up a 0.577 liter (0.577 x 10- 3 m3) glass (pyrex) beaker of cold coffee. initi

ally, the temperature is 15.8 °c, and the beaker is filled to the brim. a short time later when the student returns, the temperature has risen to 91.3 °c. the coefficient of volume expansion of coffee is the same as that of water. how much coffee (in cubic meters) has spilled out of the beaker
Physics
1 answer:
krok68 [10]3 years ago
5 0

Here coffee is filled into a glass cup and when it is heated expansion of coffee as well as glass occurs

we will find the change in volume of glass cup and change in the volume of coffee.

volume expansion is calculated by

\Delta V = V_o \gamma \Delta T

for pyrex glass we have

\gamma = 4 * 10^{-6}

now by the formula

\Delta V = 4* 10^{-6} * 0.577 * (91.3 - 15.8)

\Delta V = 0.174 mL

Now for the expansion of coffee

\gamma = 2.14 * 10^{-4}

now the expansion of coffee will be

\Delta V = 2.14 * 10^{-4}* 0.577* (91.3 - 15.8)

\Delta V = 9.32 mL

now the coffee that will spill out of the cup will be

\Delta V = 9.32 mL - 0.174 mL

\Delta V = 9.15 mL

so 9.15 mL coffee will spill out of the cup

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(a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

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Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

E=0

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}

E\propto\dfrac{1}{z^2}

(c). In terms of R,

We need to calculate the positive distance

If E\rightarrow E_{max}

Then, \dfrac{dE}{dz}=0

\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0

Taking only positive distance

z=\dfrac{R}{\sqrt{2}}

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}

E_{max}=15418.7\ N/C

E_{max}=1.54\times10^{4}\ N/C

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

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F<f

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Therefore , the farmer is not able to move the mule.

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