The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's <em>(1/18)</em> as much as on Earth.
The acceleration of gravity there would be about <em>0.545 m/s²</em>.
This is about 12% less than the gravity on Pluto.
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
Answer:
The capacitance of the deflecting plates is 
.
Explanation:
The expression for the capacitance of the capacitor in terms of area and distance is as follows;
Here, C is the capacitance, A is the area, d is the distance and 
is the absolute permittivity.
Convert the side of the square from cm to m.
s= 3.0 cm
s= 0.030 m
Calculate the area of the square.
Put s= 0.030 m.
Convert distance from mm to m.
d= 5.0 mm
Calculate the capacitance of the deflecting plates.
Put , and 
.
Therefore, the capacitance of the deflecting plates is .
