Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
For equipotential surface, dV = 0 so
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
Strength, ductility, and toughness are the three top mechanical qualities that should be given importance. Stainless steel comes with 10–30% chromium as its alloying element. It is this element that helps it resist corrosion.
Answer:
The law zero of thermodynamics.
Explanation:
The law zero of thermodynamics, which tells us that heat flows from a body at a higher temperature to another body with lower temperature, when the heat transfer is zero, it is said that the two bodies are in thermal equilibrium, their temperatures are equal
1) By looking at the table of the visible spectrum, we see that blue light has a wavelength in the range [450-490 nm], while red light has wavelength in the range [620-750 nm]. Therefore, red light has longer wavelength than blue light.
2) The frequency f of an electromagnetic wave is related to its wavelength
by the formula
where c is the speed of light. We see that the frequency is inversely proportional to the wavelength, so the shorter the wavelength, the greater the frequency. In this case, blue light has shorter wavelength than red light, so blue light has greater frequency than red light.
3) The energy of the photons of an electromagnetic wave is given by
where h is the Planck constant and f is the frequency. We see that the energy is directly proportional to the frequency, so the greater the frequency, the greater the energy. In this problem, blue light has greater frequency than red light, so blue light has also greater energy than red light.
Answer:
Pressure, 
Explanation:
It is given that,
Mass of the woman, m = 55 kg
Diameter of the circular cross section, d = 6 mm
Radius, r = 3 mm = 0.003 m
Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :
So, the pressure exerted on the floor is 
. Hence, this is the required solution.
