Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :
So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As 'k' is coulomb's constant it's value is 9 x N m²/C²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]