Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

Answer:

The work done, W = 19.6 J

Explanation:

It’s provided that

The mass of the block, m = 5 kg

The velocity of the block, v = 10 m/s

The coefficient of kinetic friction between the block and rough surface is 0.2

Distance traveled by the block, d = 2 m

As the block traverses the rough section, it loses energy equal to the work done by the kinetic energy.

W = 19.6 J

Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.

Answer:

50.2 cm

Explanation:

We have the following data:

Height, h=3.5 m

Initial horizontal velocity,

Time, t=0.32 s

We need to determine how far the ball is from the ground after 0.32 s.

Initial vertical velocity,

Where

Answer:

The particle's energy in its ground state is E₁=1.5 eV.

Explanation:

For a particle with mass m in the nth energy level of an infinite square well potential of width L , the energy is given by:

                                                   

In the ground state (n=1) and in the first excited state (n=2), where the energy is noted as E₂= 6.0 eV. Substituting into the above equation yields:

                                                               

                                                   

Thus, we can express the ground state's energy as:

                                                   

                                                     

                                                   

Ultimately

                                                   

                                                   

                                                   

The terminal velocity of the object is 12.58 m/s. Explanation: Terminal velocity occurs when the drag force equals the weight of the object. The gravitational force is calculated as mg = 80 * 9.8 = 784 N. The drag force can then be equated to solve for the terminal velocity, resulting in v = 12.58 m/s or v = -15.58 m/s (which is unfeasible). Thus, the terminal velocity of the falling object is 12.58 m/s.