Physics, 23.09.2019 20:30 myohmyohmy

Suppose mary would like to design an emergency overheating alarm using one of these materials. in her design, at room temperature (23.0 ∘c)(23.0 ∘c) , a 6.01 cm6.01 cm long cylinder of the material will sit with one end against a wall and the other just barely touching a button. once the heat causes the cylinder to lengthen by 0.0273 cm0.0273 cm , the button will be activated, turning on the alarm. a) at what temperature will the alarm be triggered if the cylinder is made out of polycarbonate? degrees c
b) at what temperature will the alarm be triggered if the cylinder is made out of cast iron? degrees c

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Answer from: loistaylor1819

A. 92.88 °C

B. 401.535 °C

Explanation:

So, the overheating system is going to be based on the principle of thermal linear expansion. The behavior of this principle is ruled by the equation:

∆L= L_i * ∆T * α

Where α is a coefficient of linear expansion. For a cylinder made from polycarbonate α = 70,2*10^(-6) °C^(-1) and for a cylinder made from cast iron α = 12*10^(-6) °C^(-1). If we isolate the term of the temperature’s difference, we have:

∆L/(L_i * α) = ∆T → T_f = T_i + ∆L/(L_i * α)

Replacing the values, for the case of the Polycarbonate we have:

T_f = T_i + ∆L/(L_i * α) = 23°C+0,0273cm/(6,01cm *70,2 * 10^(-6)°C^(-1) ) = 92,88 °C

Replacing the values, for the case of the Cast Iron we have:

T_f = T_i +∆L/(L_i * α) = 23°C + 0,0273cm/(6,01cm * 12 * 10^(-6) °C^(-1) ) = 401,535 °C

As we see, is way better to use the polycarbonate in this application.

Have a nice day. Let me know if I can help you with anything else. :D

Answer from: Quest

Ireally don’t know

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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