Answer:
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
between A and E. (You will need a protractor to mcasure angles.) 1. Place the loop over the nails so that the number (# of field lines through surface) of field lines through it is a maximum, Determine the angle between A and E. Record both that angle and the number of field lines that pass through the loop. 2. Rotate the loop until there is one fewer row of nails passing through it. Determine the angle between A and E and record your measurement. Continue in this way until θ= 180. 3. On graph paper, plot a graph of n versus o (Let the number of field lines through the surface be a negative number for angles between 90 and 180.) When E and A were parallel, we called the quantity EA the electric flux through the surface. For the parallel case, we found that EA is proportional to the number of field lines through the surface. E. By what trigonometric function of θ must you multiply EA so that the product is proportional to the number of ficld lines through the area for any orientation of the surface? Rewrite the quantity described above as a product of just the vectors E and A.
1 answer:
Answer:
a. cosθ b. E.A
Explanation:
a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function
b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A
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Answer:
The energy coverted to heat is 200 kilojoules.
Explanation:
GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:
Where:
,
- Masses of the small meteorite and the communication satellite, measured in kilograms.
,
- Speeds of the small meteorite and the communication satellite, measured in meters per second.
- Final speed of the satellite-meteorite system, measured in meters per second.
The final speed of the satellite-meteorite system is cleared:
If ,
,
and
, the final speed is now calculated:
Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:
Where:
,
- Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.
- Kinetic energy of the satellite-meteorite system, measured in joules.
- Dissipated heat, measured in joules.
The previous expression is expanded by using the definition for the translational kinetic energy:
Given that ,
,
,
and
, the dissipated heat is:
The energy coverted to heat is 200 kilojoules.