Answer:
a. Acceleration = 1.75 m/s²
b. Force = 105 Newton.
Explanation:
<u>Given the following data:</u>
Mass = 60kg
Final velocity = 7m/s
Time = 4 seconds
Required to find the acceleration and force of the body;
a. To find the acceleration;
Mathematically, acceleration is given by the equation;
Where;
- a is acceleration measured in
- v and u is final and initial velocity respectively, measured in
- t is time measured in seconds.
Since the elevator is starting from rest, its initial velocity is equal to 0m/s.
Substituting into the equation, we have;
<em>Acceleration = 1.75 m/s²</em>
b. To find the force;
Substituting into the equation, we have;
<em>Force = 105 Newton. </em>
Answer:
q2 = -4.35*10^-9C
Explanation:
In order to find the values of the second charge, you use the following formula:
(1)
V: electric potential = 1.14 kV = 1.14*10^3 kV
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1: charge 1 = 8.60*10^-9 C
q2: charge 2 = ?
r1: distance to the first charge = 20.7mm = 20.7*10^-3 m
r2: distance to the second charge = 15.1mm
You solve the equation (1) for q2, and replace the values of the other parameters:
![q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C](https://tex.z-dn.net/?f=q_2%3D%5Cfrac%7Br_2%7D%7Bk%7D%5BV-k%5Cfrac%7Bq_1%7D%7Br_1%7D%5D%3D%5Cfrac%7BVr_2%7D%7Bk%7D-%5Cfrac%7Bq_1r_2%7D%7Br_1%7D%5C%5C%5C%5Cq_2%3D%5Cfrac%7B%281.14%2A10%5E3V%29%2815.1%2A10%5E%7B-3%7Dm%29%7D%7B8.98%2A10%5E9Nm%5E2%2FC%5E2%7D-%5Cfrac%7B%288.60%2A10%5E%7B-9%7DC%29%2815.1%2A10%5E%7B-3%7Dm%29%7D%7B20.7%2A10%5E%7B-3%7Dm%7D%5C%5C%5C%5Cq_2%3D-4.35%2A10%5E%7B-9%7DC)
The values of the second charge is -4.35*10^-9C
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
Know recall that:
We now need to know how much distance is traveled during a minute or 60 seconds:
Finally we divide this result with t=60 seconds:
Where the distance units were named units as the length unit is not specified in this exercise.<span />
