A particle of charge +18.5C and mass 4.84 × 10-5 kg is released from rest in a region where there is a constant electric field of +578 N/C. What is the displacement of the particle after a time of 1.89 × 10-2 s?
1) F=Eq
2) a=F/m
3) d=vt+1/2at^2
1) F= 10693 N
2) a=220929752 m/s^2
3) d= 39459.15837 m
Velocity is 0 m/s bec it was released at rest.
I substituted Equation 1 (F) in 2 and then 2 (a) in 3.
These are the equations I used to get my answer but the homework tells me I am wrong.
It seems like you have correctly applied the relevant equations to solve the problem. However, let's go through the calculations step by step to make sure everything is correct.
1) First, we can use the equation F = Eq to find the force acting on the particle:
F = (+18.5 C) * (+578 N/C) = 10667 N (rounded to the nearest whole number)
2) Next, we can use the equation a = F/m to find the acceleration of the particle:
a = (10667 N) / (4.84 × 10^-5 kg) = 2.2 × 10^8 m/s^2 (rounded to the appropriate number of significant figures)
3) Finally, we can use the equation d = vt + 1/2at^2 to find the displacement of the particle. Since it was released from rest, its initial velocity is 0 m/s:
d = (0 m/s) * (1.89 × 10^-2 s) + 1/2 * (2.2 × 10^8 m/s^2) * (1.89 × 10^-2 s)^2
= 0 + 1/2 * (2.2 × 10^8 m/s^2) * (3.5721 × 10^-4 s^2)
= 1/2 * (7.86442 × 10^4) = 3.93221 × 10^4 m (rounded to the appropriate number of significant figures)
Therefore, the displacement of the particle after a time of 1.89 × 10^-2 s is approximately 3.93221 × 10^4 m.
If your homework is telling you that your answer is wrong, double-check your calculations and make sure you entered the values correctly. Additionally, check if there are any additional details or conditions given in the problem statement that you may have missed.