Answer:
The magnitude of the force required to move the electron through the given field is 2.203 N
Explanation:
Given;
The field strength of the electron, E = 1.375 x 10¹⁹ N/C
charge of electron, q = 1.602 x 10⁻¹⁹ C
The magnitude of the force required to move the electron through the given field is calculated as follows;
F = Eq
F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)
F = 2.203 N
Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N
-- volume = (length)(width)(height)
-- Since the cube is a cube, its three dimensions are all the same number.
Volume = (2.5cm)(2.5cm)(2.5cm)
Volume = 15.625 cubic cm
-- density = (mass) / (volume)
Density = (1129.56g) / (15.625cm^3)
Density = 72.3 g/cm^3
(roughly 3.2 TIMES the density of the most dense naturally occurring substance on Earth)
The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass, 
= 72 kg
speed, 
= 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X + X = ( + ) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X 
v = 417.6 + 
v = 5.87 m/s
The final velocity is 5.87 m/s
Answer:
im pretty sure its c the third answer i got that one right
Explanation:
your welcome :)
