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Elis [28]
3 years ago
14

Breaking a pencil in half is an example of: options:

Physics
2 answers:
Licemer1 [7]3 years ago
8 0

Answer:

b

Explanation:

IRISSAK [1]3 years ago
8 0
Answer
physical change

explanation
breaking the pencil is only changing the physical form and you can always sharpen it to fix it

hope this helps and have a wonderful day :)
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What is a cool example of newtons 3rd law?
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What is the Net Force?
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3 years ago
Compare the results of applying the acceleration equation in the following 2 cases : (1) an object that goes from 0 to 10 m/s in
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3 years ago
A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1
nasty-shy [4]

(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

R=\rho \frac{L}{A} (1)

where

\rho is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m, so the cross-sectional area is

A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega

And now we can re-arrange the eq.(1) to solve for the resistivity:

\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m

(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

where

R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

T=92^{\circ}C

T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

5 0
3 years ago
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