The shell which can have maximum of 8 electrons.

1) the correct answer is B ,  2)  θ= 45º,

3) the projectile A must have a greater angle ,

4A)   vₓ (B)> vₓ (A) ,  

5B)  The maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

Explanation:

In this long exercise you are asked to answer a series of questions about the launching of projectiles

1) Path type

since they indicate that the air resistance is negligible the horizontal displacement is

       x = v₀ₓ t

On the vertical axis, a relationship acts, which is gravity, therefore its displacement is

      y = t - ½ g t²

let's substitute to eliminate time

     y = v_{oy} (x / v₀ₓ) - ½ g (x / v₀ₓ)²

    y = (tan θ) x - (g / v₀ₓ²) x²

we can see that this is the equation of a parabola

therefore the correct answer is B

2) From the missile launch expressions the range is

           R = x = v₀² sin² 2θ / g

for the range to be maximum sin 2θ = 1, this occurs for

          2θ = 90

         θ= 45º

the correct answer is "The projectile fired at an angle close to 45º lands farther"

In response D implies that the projectile has been divided into two parts, in this case it is true that the lighter part must go further.

3) From the expression of scope we can see that as the angle increases the square sine becomes smaller, for example without 180 = 0

therefore to reach the nearest ship A must launch with a greater angle, therefore the projectile A must have a greater angle

Respect the questions are a bit strange.

* If the projectiles are fired at the same angle and in the statement they indicate that the velocities are equal, the two must reach the same height 4

4A. If the vertical speed is equal, the projectile that must reach the furthest ship (B) must have more horizontal speed

          vₓ (B)> vₓ (A)

5B. Both are launched with the same horizontal speed.

which is higher

            ² =   v_{oy}² - 2 g y

at maximum height vertical velocity is zero (v_{y} = 0)

            0 = v_{oy}² - 2 g y

            y = v_{oy}² / 2g

the maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) they both reach the same height

As the value of the acceleration due to gravity is constant, the time to go up is the same time to go down, therefore if the two projectiles reach the maximum height, the time used is

              = v_{oy} - g t

              0 = v_{oy} - gt

              t = v_{oy} / g

if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

1) the correct answer is B ,  2)  θ= 45º,

3) the projectile A must have a greater angle ,

4A)   vₓ (B)> vₓ (A) ,  

5B)  The maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

Explanation:

In this long exercise you are asked to answer a series of questions about the launching of projectiles

1) Path type

since they indicate that the air resistance is negligible the horizontal displacement is

       x = v₀ₓ t

On the vertical axis, a relationship acts, which is gravity, therefore its displacement is

      y = t - ½ g t²

let's substitute to eliminate time

     y = v_{oy} (x / v₀ₓ) - ½ g (x / v₀ₓ)²

    y = (tan θ) x - (g / v₀ₓ²) x²

we can see that this is the equation of a parabola

therefore the correct answer is B

2) From the missile launch expressions the range is

           R = x = v₀² sin² 2θ / g

for the range to be maximum sin 2θ = 1, this occurs for

          2θ = 90

         θ= 45º

the correct answer is "The projectile fired at an angle close to 45º lands farther"

In response D implies that the projectile has been divided into two parts, in this case it is true that the lighter part must go further.

3) From the expression of scope we can see that as the angle increases the square sine becomes smaller, for example without 180 = 0

therefore to reach the nearest ship A must launch with a greater angle, therefore the projectile A must have a greater angle

Respect the questions are a bit strange.

* If the projectiles are fired at the same angle and in the statement they indicate that the velocities are equal, the two must reach the same height 4

4A. If the vertical speed is equal, the projectile that must reach the furthest ship (B) must have more horizontal speed

          vₓ (B)> vₓ (A)

5B. Both are launched with the same horizontal speed.

which is higher

            ² =   v_{oy}² - 2 g y

at maximum height vertical velocity is zero (v_{y} = 0)

            0 = v_{oy}² - 2 g y

            y = v_{oy}² / 2g

the maximum height is reached by the projectile with the highest speed, from the front this is projectile B

7) they both reach the same height

As the value of the acceleration due to gravity is constant, the time to go up is the same time to go down, therefore if the two projectiles reach the maximum height, the time used is

              = v_{oy} - g t

              0 = v_{oy} - gt

              t = v_{oy} / g

if the vertical velocity of both projectiles is the same time is the same

therefore the two projectiles have equal flight times

A

Explanation:

The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it useful for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with  s2p6 .

A

Explanation:

The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it useful for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with  s2p6 .

The answer is below

Explanation:

a) The volume of a sphere is:

Volume = (4/3)πr³; where r is the radius of the shell.

Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:

Volume of spherical shell = cm³

But Density = mass / volume;   Mass = density * volume.

Therefore, mass of spherical shell = density * volume

mass of spherical shell = * cm³

Mass of liquid = volume of inner shell * density of liquid

Mass of liquid =

Total mass of sphere including contents = mass of spherical shell + mass of liquid

Total mass of sphere including contents (M) = *  +   =

Total mass of sphere including contents (M) = (346 - 14.5a³) grams

b) The mass is maximum when the value of a = 0

M = 346 - 14.5a³

M' = 43.5a² = 0

43.5a² = 0

a = 0

The answer is below

Explanation:

a) The volume of a sphere is:

Volume = (4/3)πr³; where r is the radius of the shell.

Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:

Volume of spherical shell = cm³

But Density = mass / volume;   Mass = density * volume.

Therefore, mass of spherical shell = density * volume

mass of spherical shell = * cm³

Mass of liquid = volume of inner shell * density of liquid

Mass of liquid =

Total mass of sphere including contents = mass of spherical shell + mass of liquid

Total mass of sphere including contents (M) = *  +   =

Total mass of sphere including contents (M) = (346 - 14.5a³) grams

b) The mass is maximum when the value of a = 0

M = 346 - 14.5a³

M' = 43.5a² = 0

43.5a² = 0

a = 0

Following are the code to this question:

Counter T = new Counter(1, 100); //creating Counter class object and call its parameterized constructor        

Thread T1 = new Thread(new Runnable()  //creating Thread object T1.

{            

   public void run() //define run method

   {                

       T.countRange(); //call countRange method    

   }          

});    

Thread T2 = new Thread(new Runnable()  //creating another object "T2" of Thread.

{            

   public void run() //define run method

   {                

       T.countRange();           //call countRange method

   }      

});    

T1.start(); //start Thread T1    

T2.start(); //start Thread T2    

Explanation:

Description of the above code as follows:

Following are the code to this question:

Counter T = new Counter(1, 100); //creating Counter class object and call its parameterized constructor        

Thread T1 = new Thread(new Runnable()  //creating Thread object T1.

{            

   public void run() //define run method

   {                

       T.countRange(); //call countRange method    

   }          

});    

Thread T2 = new Thread(new Runnable()  //creating another object "T2" of Thread.

{            

   public void run() //define run method

   {                

       T.countRange();           //call countRange method

   }      

});    

T1.start(); //start Thread T1    

T2.start(); //start Thread T2    

Explanation:

Description of the above code as follows: