A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  

Explanation:

The question indicates that

    The wire’s diameter is  

     The radius of the wire is  

     Aluminum's resistivity is

       The electric field variation is described as

The charge is effectively given by the equation

Where A is the area expressed as

 Thus,

Therefore

By substituting values

The question states that t =  5 seconds

Answer:

Explanation:

The transitions occur as follows:

P₁ V₁ changes to 3P₁, V₁ (with constant volume) — first phase.

Subsequently, 3P₁,V₁ transitions to 3P₁, 5V₁ (with constant pressure) — second phase.

During the initial phase, the temperature must be escalated by a factor of 3. Therefore, if the starting temperature is T₁, then the ending temperature will be 3 T₁.

P₁V₁ = n R T₁, where n represents the number of moles of gas.

Thus, nRT₁ = P₁V₁.

The heat added at constant volume is given by n Cv (3T₁ - T₁),

= n x 5/3 R X 2T₁ (noting that for diatomic gas, Cv = 5/3 R).

= 10/3 x nRT₁

= 10/3 x P₁V₁.

In the second phase, the temperature must rise 5 times. Thus, if the initial temperature is 3T₁, then the final temperature will be 15 T₁.

The heat added at constant pressure in this scenario becomes:

= n Cp (15T₁ - 3T₁)

= n x 7/3 R X 12T₁ (for diatomic gases, Cp = 7/3 R).

= 28 x nRT₁

= 28 P₁V₁.