<span>let r be side of square
V (B) = k q2/r + k Q / sqrt[r^2 + r^2] = 0 given
q2/r + Q / r root2 = 0
q2 = - Q / root2
q2/Q = - 1/root2
q2/Q = - root2 /2 = - 0.707
q2 equals negative 0.707 times Q</span>
Answer: Given:
Initial velocity= 36km/h=36x5/18=10m/s
Final velocity =54km/h=54x5/18=15m/s
Time =10sec
Acceleration = v-u/ t
=15-10/10=5/10=1/2=0.5 m/s2
Distance =s=?
From second equation of motion:
S=ut +1/2 at^2
=10*10+1/2*0.5*10*10
=100+25
=125m
So distance travelled 125m
Hope it helps you
This is defiantly not from Elementary School.
Answer:
7.6427m/s
Explanation:
Given:![v_a=4.7m/s, \theta_a=33.0\textdegree and \ v_b=4.5m/s](https://tex.z-dn.net/?f=v_a%3D4.7m%2Fs%2C%20%5Ctheta_a%3D33.0%5Ctextdegree%20and%20%5C%20v_b%3D4.5m%2Fs)
#Applying the conservation of momentum along the x-axis:![mv_i=mv_acos\theta_a+mv_bcos\theta_b](https://tex.z-dn.net/?f=mv_i%3Dmv_acos%5Ctheta_a%2Bmv_bcos%5Ctheta_b)
#And along y-axis:
![0=-sin\theta_a+mv_bsin\theta_b](https://tex.z-dn.net/?f=0%3D-sin%5Ctheta_a%2Bmv_bsin%5Ctheta_b)
#Solving for
:
![sin\rheta_b=\frac{v_a}{v_b}sin\theta_a=4.7/4.5\times sin33.0\textdegree\\=0.5688\\\therefore \theta_b=34.67\textdegree](https://tex.z-dn.net/?f=sin%5Crheta_b%3D%5Cfrac%7Bv_a%7D%7Bv_b%7Dsin%5Ctheta_a%3D4.7%2F4.5%5Ctimes%20sin33.0%5Ctextdegree%5C%5C%3D0.5688%5C%5C%5Ctherefore%20%5Ctheta_b%3D34.67%5Ctextdegree)
#By substitution in the x-axis equation:
![v_i=4.7cos 33.0\textdegree +4.5cos 34.67\textdegree\\=7.6427m/s](https://tex.z-dn.net/?f=v_i%3D4.7cos%2033.0%5Ctextdegree%20%2B4.5cos%2034.67%5Ctextdegree%5C%5C%3D7.6427m%2Fs)
Hence the original speed of the ball before impact is 7.6427m/s
Answer:
You'd have to show the picture..