Answer:
The value is 
Explanation:
From the question we are told that
The velocity of the each of the three cars is 
The velocity of the fourth car is 
The initial velocity of the fifth car 
Generally from the law of momentum conservation we have that
![m_1 u_1 + m_2 u_2 + m_3 u_3 +m_4u_4 + m_5u_5 = [m_1 + m_2 + m_3 +m_4+ m_5]v](https://tex.z-dn.net/?f=m_1%20u_1%20%2B%20m_2%20u_2%20%2B%20m_3%20u_3%20%2Bm_4u_4%20%2B%20m_5u_5%20%3D%20%20%5Bm_1%20%20%20%2B%20m_2%20%2B%20m_3%20%2Bm_4%2B%20m_5%5Dv)
Given that the cars are identical then their mass will be the same
i.e

=> ![[u_1 + u_2 + u_3 +u_4 + u_5]m = 5mv](https://tex.z-dn.net/?f=%5Bu_1%20%2B%20u_2%20%2B%20%20u_3%20%2Bu_4%20%2B%20u_5%5Dm%20%3D%20%205mv)
=> 
= > 
Answer:
a) 096V b) 0.0288A c) 0.3456W
Explanation:
a) Vp/Vs= Np/Ns
120/Vs= 500/4
Vs= 096V
b) Np/Ns= Is/Ip
500/4= 3.6/Ip
Ip= 0.0288A
c) P= VI
P=(120)(0.0288)
P= 0.3456W
Answer:
Zero
Explanation:
here, the inductive reactance and the capacitive reactance is same, so this is the condition for resonance.
In the condition for resonance,
the circuit and the voltage in the circuit is in the same phase and the impedance of the circuit is minimum which is equal to the resistance of the circuit.
The phase angle is given by

Ф = 0
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D-sin%28-%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%2Bsin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
Since 
where;
L = length
Q = charge
λ = density of the charge;
then substituting
for λ, we have :
![E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%28%5Cfrac%7BQ%7D%7BL%7D%29%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
![E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2Q%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20E_oLR%7D)
substituting our given parameter; we have:
![E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%286.26%2A10%5E%7B-12%7DC%29%5Bsin%5Cfrac%7B2.59rad%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FN.m%5E2%29%280.235431%29%280.0909%29%7D)
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C