Answer:
0.8 N
Explanation:
From coulomb's law,
Formula:
F = kqq'/r²........................ Equation 1
Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.
Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 1
F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²
F = 1800×10⁻³/2.25
F = 1.8/2.25
F = 0.8 N
<span>Quarks are thought to be the basic component of protons and newtons.</span>
Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
Answer:
![F_a_v_g=7093333.33N*s](https://tex.z-dn.net/?f=F_a_v_g%3D7093333.33N%2As)
Explanation:
The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:
![F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}](https://tex.z-dn.net/?f=F_a_v_g%3Dm%2A%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%3Dm%2A%5Cfrac%7Bv_2-v_1%7D%7Bt_2-t_1%7D)
Where:
![m=mass\hspace{3}of\hspace{3}the\hspace{3}object](https://tex.z-dn.net/?f=m%3Dmass%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject)
![v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval](https://tex.z-dn.net/?f=v_2%3Dfinal%5Chspace%7B3%7Dvelocity%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject%5Chspace%7B3%7Dat%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dend%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dtime%5Chspace%7B3%7Dinterval)
![v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.](https://tex.z-dn.net/?f=v_1%3Dinitial%5Chspace%7B3%7Dvelocity%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject%5Chspace%7B3%7Dwhen%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dtime%5Chspace%7B3%7Dinterval%5Chspace%7B3%7Dbegins.)
![t_2=final\hspace{3}time](https://tex.z-dn.net/?f=t_2%3Dfinal%5Chspace%7B3%7Dtime)
![t_1=initial\hspace{3}time](https://tex.z-dn.net/?f=t_1%3Dinitial%5Chspace%7B3%7Dtime)
Asumming v1=0 and t1=0:
![F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s](https://tex.z-dn.net/?f=F_a_v_g%3Dm%2A%20%5Cfrac%7Bv_2%7D%7Bt_2%7D%20%3D%2824.7%29%2A%5Cfrac%7B784%7D%7B2.73%2A10%5E%7B%2A3%7D%20%7D%20%3D7093333.333N%2As)
Answer: a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Explanation:
Acceleration is the rate of change in the velocity per time
a = change in velocity/time
a = ∆v/t
average acceleration a = (v2 -v1)/t. ....1
Given;
Final velocity v2 = 1.63m/s
Initial velocity v1 = -1.15ms
time taken t = 2.11s
Substituting into eqn 1
a = [1.63 - (-1.15)]/2.11
a = (1.63+1.15)/2.11
a = 2.78/2.11
a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2