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hram777 [196]
3 years ago
6

Suppose Mary would like to design an emergency overheating alarm using one of these materials. In her design, at room temperatur

e (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
Physics
1 answer:
Reptile [31]3 years ago
4 0

Answer:

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

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3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

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and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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