Answer:
a) [tex]E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) zero
Explanation:
a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:
[tex]E_T=E_1+E_2[/tex]
E1: electric field of charge 1
E2: electric field of charge 2
It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:
[tex]E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\[/tex]
r13: distance between charges 1 and 3
r12: charge between charges 2 and 3
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
Thus, you first calculate the distance r13 and r23, and also the angles:
[tex]r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°[/tex]
Next, you replace the values of all parameters in order to calculate E1 and E2:
[tex]E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}[/tex]
finally, you obtain for ET:
[tex]E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.
A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,
find the velocity of bullet.
Answer:
-450 m/s
Explanation:
Momentum is conserved.
p₀ = p
0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v
v = -450 m/s