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
I believe the correct answer from the choices listed above is option B. The sun is a main sequence type of star. <span>The sun is classified as a G2V star, sometimes referred to as a yellow dwarf. It is a Population I star in its main </span>sequence<span>. Hope this answers the question.</span>
Answer:
jk
Explanation:
Lets decompose the initial velocity into its components:
Vi = 2.25 m/s
Vix = Vi x cos(50)
Viy = Vi x sin(50) = 2.25 x sin(50) = 1.724
Then decompose the final velocity:
Vf = 4.65
Vfx = Vf x cos(120)
Vfy = Vf x sin(120) = 4.65 x sin(120) = 4.027
After that we can use:
Vfy = Viy + ay*t
ay = (Vfy - Viy)/t
ay = (4.027 - 1.724) / 8.33
ay = 0.276
Acceleration of an object is depended upon the net force acting open the object and the mass of the object
