Answer:
x =4.5 10⁴ m
Explanation:
To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law
m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg
q = -200 nc (1C / 10 9 nC) = -200 10-9 C
Let's calculate the acceleration
F = ma
F = q E
a = qE / m
a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶
a = 1 10² m / s²
Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)
Vf² = Vo² -2 a x
0 = Vo² - 2 a x
x = Vo² / 2a
x = 3000²/ 2100
x =4.5 10⁴ m
This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial
Second part
In this case Newton's second law is applied on the y axis
F -W = 0
F = w = mg
E q = mg
E = mg / q
E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹
E = 9.8 10⁵ C
The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E
The correct answer is false.
1) Time to get the highest point of it pathway>
Vf = Vo - g*t
Vf = 0 (condition of maximum height)
Vo = 5.33 m/s
g = 9.81 m/s
=> t = Vo / g = 5.33 m/s / 9.81 m/s = 0.543 s
2) Time to return from the highest point = time to get to the highest point = 0.543 s
Total time = 0.543s + 0.543 s = 1.086 s ≈ 1.09 s
Answer: 1.09 s
Answer: concentration c = 25.2 g /(41.99 g/mol × 0.75 l)
Explanation: M(NaF) = 41.99 g/mol and amount of substance n = m/M.
Concentration c = n/V
