Ability of the muscles to function effectively and efficiently without undue fatigue

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.

Answer:

The radius is

Explanation:

The problem states that

    The magnetic field is  

     The electron kinetic energy is  

In general, for a collision to happen, the centripetal force on the electron in its orbit must equal the magnetic force acting on it  

   This can be mathematically expressed as

   

=>    

Where  m denotes the electron’s mass, which has a value of  

             v signifies the escape velocity, mathematically represented as

               

Thus,

       

     applying indices

   

substituting these values

   

       

       

     

To determine the average velocity of an object descending in a vacuum, assuming you know its final speed, multiply the final result by the overall time. 3. The equation d = v • t expresses how distance, average velocity, and time relate to each other.

Answer:

x = 0.29 m

Explanation:

It is known that the total external force acting on the mass system equals ZERO,

so the center of mass of the entire system will stay stationary.

We find that

Since Ernie approaches Burt's position, we have:

therefore, we conclude that

Answer:

The area that remains is 4.201 m²

Explanation:

Provided that

AB=D=  4 m  (R=2 m)

The area of the half-circle AB is

A=2π

The area of the smaller half-circle is

a=5π/6 + 2√3/4  m²

a=5π/6 + √3/2  m²

<pThus, the remaining area = A - a

                                     = 2π - (5π/6 + √3/2) m²

The remaining area is expressed as  2π - (5π/6 + √3/2) m²