Why do scientist use different types of models to represent compounds

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

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris

Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_{x} = 0$

$T + F_{AB} Sin 60$ = 0 ...... (1)

$\sum F_{y}$ = 0

$F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_{allow} = \frac{T}{A}$

T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

= 3925 kip

From equation (1), $F_{AB}Sin (60^{o})$ = -3925

$F_{AB} \times -0.304 8$ = -3925

$F_{AB}$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip \times \frac{1}{2} + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0

3 years ago

Compare and contrast the energy transfer of a roller coaster to that of a pendulum

Softa [21]

When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

<h3>Compare and contrast the energy transfer of a roller coaster to that of a pendulum:</h3><h3>What is the transfer of energy in a roller coaster?</h3>

The transfer of potential energy to kinetic energy occur when the roller coaster move along the track. As the motor pulls the cars to the top, the body has more potential energy whereas when the body comes to the bottom , it has kinetic energy in the object.

<h3>What is the energy transfer in a pendulum?</h3>

As a pendulum swings, its potential energy changes to kinetic energy and kinetic energy changes into potential energy. At the top more potential energy is present.

So we can conclude that When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

Learn more about energy here: brainly.com/question/13881533

#SPJ1

8 0

2 years ago

A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet

Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

$\varepsilon =-\frac{d\phi_B}{dt}$

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

$V=IR$

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0

3 years ago

Read 2 more answers

PLEASE HELP AND HURRY

andreyandreev [35.5K]

C
I took the tests earlier hope this helps

7 0

4 years ago