A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of

Answer:

A) The updated amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please refer to the attached documents for the solution.

Answer:

At this position, the magnetic field equals ZERO

Explanation:

The magnetic field produced by a moving charge is described as

Here, we determine the direction of the magnetic field using

Thus, we find

Leading to a magnetic field of ZERO

Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent

The change in momentum (i.e., impulse) from the car during the collision remains constant regardless of whether an airbag is present. This is because the car's mass is unchanged, and the velocity change remains the same. Therefore, if the force is constant as F and reduced by a factor of 110, it follows that the collision duration must increase by the same factor when the airbag is utilized.

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass

Radius

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

Where G represents the gravitational constant and M denotes the mass of the planet.

Now, let’s compute the time value.

Ultimately, the time for the rock to hit the surface is t = 87.58s.

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

Pressure at point (1) P1 = 10 bar = P2

Volume at point (1) V1 = 1 m^3

Volume at point (2) V2 =4 m^3

For Process 2 ---> 3, where V = constant

Pressure at point (3) P3 = 10 bar

Volume at point (3) V3 = 4 m^3

Process 3 ---> 1 defined as PV = constant.

Required

Sketch the processes on the PV coordinates To calculate the work in kJ

Work is calculated by W=

a=V2

b=V1

x=Pdx=dV

For Process 1 ---> 2 where P3 = P4 = 5 bar  

W=

a=V3

b=V2

x=4dx=dV

substituting the values here into the integral gives

W=400 kJFor Process 2 ---> 3

As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ  

For Process 3 ---> 1  By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1  

 W=

a=V1

b=V3

x=1V^-1dx=dVsubstituting values into this integral results in

W=| ln V | limits a and b

  = -160.944 KJ