Answer:
B
Explanation:
Given:-
- The charge of the test particle q = 3.0 * 10^-9 C
- The force exerted by the metal sphere F = 6.0 * 10^-5 N
Find:-
The magnitude and direction of the electric field
strength at this location?
Solution:-
- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:
F = E*q
- Using the data given we can determine E:
E = F / q
E = (6.0 * 10^-5) / (3.0 * 10^-9)
E = 20,000 N/C
- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.
Answer:
Day and night on Earth are equal.
Explanation:
Vernal means fresh or new like the spring. The vernal equinox, because it signals the beginning of spring. On this day, the Sun is exactly above the Equator and day and night are of equal length. In the northern hemisphere, the vernal equinox is around March 20 or 21 when the Sun crosses the celestial equator going north while in the southern hemisphere, the vernal equinox is around September 22 or 23 when the Sun moves south across the celestial equator.
ONE CAN perform this by doing an ideal experiment
by creating an isothermal system
its like you supply heat to a body and that body is present at very low temperature the amount of heat you supply is equal to the amount of heat lost by that body due to difference in the temperature of the body and the surrounding. heating curve will be constant as there is no change in the internal energy of the system ..
Answer:
Average speed = 8.8 m/s
Average velocity = 8.8 m/s
Explanation:
From the question,
Avearge speed = 2πr/t................ Equation 1
Where r = 14 m, t = 10 s
Constant π = 22/7
Substitute this value into equation 1
Average speed = 2(22/7)(14)/10
Avearge speed = 8.8 m/s.
For a body moving in circular motion, The average speed and average velocity are equal.
Hence,
Average speed = 8.8 m/s
Average velocity = 8.8 m/s
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=
as E=
so V=
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=
q=27× coulomb
=3.6× coulomb [ans]