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3 years ago

Study the roller coaster diagram. Match the labels to the correct location on the roller coaster.

Physics

2 answers:

Leni [432]3 years ago

5 0

Answer:

A: Greatest Potential Energy B: Increasing Potential Energy C: Greatest Kinetic Energy D: Increasing Kinetic Energy

Explanation:

Hope this helps and have a good day!

IRISSAK [1]3 years ago

3 0

Answer:

A- Greatest Kinetic Energy

B- Increasing Potential Energy

C- Increasing Kinetic Energy

D- Greatest Potential Energy

Explanation:

hope this is right.

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Suspend a heavy weight by two pieces of rope. Tension is greatest when the ropes

bija089 [108]

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4 years ago

Which of the following is the correct definition of conduction? A. the transfer of heat by currents within a liquid or gas B. th

dolphi86 [110]

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3 years ago

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In a transpiration experiment, the air bubble has an initial volume of 4.33 mL, and an initial pressure of 101.9 kPa. What will

ira [324]

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3 years ago

-Which of the following wires will have the least resistance?

scoray [572]

The wires that will have the least resistance is :
C. A short thick wire
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7 0

3 years ago

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When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length

monitta

Answer:

$3.34\Omega$

Explanation:

The resistance of a metal rod is given by

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity

L is the length of the rod

A is the cross-sectional area

The resistivity changes with the temperature as:

$\rho(T)=\rho_0 (1+\alpha (T-T_0))$

where in this case:

$\rho_0$ is the resistivity of silver at $T_0=21.0^{\circ}C$

$\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}$ is the temperature coefficient for silver

$T=180.0^{\circ}C$ is the current temperature

Substituting,

$\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0$

The length of the rod changes as

$L(T)=L_0 (1+\alpha_L(T-T_0))$

where:

$L_0$ is the initial length at $21.0^{\circ}C$

$\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}$ is the coefficient of linear expansion

Substituting,

$L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0$

The cross-sectional area of the rod changes as

$A(T)=A_0(1+2\alpha_L(T-T_0))$

So, substituting,

$A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0$

Therefore, if the initial resistance at 21.0°C is

$R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega$

Then the resistance at 180.0°C is:

$R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega$

7 0

3 years ago