Menu Close

Question-153959




Question Number 153959 by DELETED last updated on 12/Sep/21
Answered by DELETED last updated on 12/Sep/21
F_(AC) =k.((q_a .q_c )/r_(AC) ^2 )           =9.10^9 .((2.10^(−6) .2.10^(−4) )/((30.10^(−2) )^2 ))        =((36)/9)×10^(9−6−4+4−2)        =4×10^1 =40 N  F_(BC) =k.((q_b .q_c )/r_(BC) ^2 )     =9.10^9 .((2.10^(−6) .2.10^(−4) )/((30.10^(−2) )^2 ))     =((36)/9)×10^(9−6−4+4−2)         =4×10^1 =40 N  F_R =(√(F_(ac) ^2 +F_(bc) ^2 +2F_(ac) .F_(bc) cos 60°))       =(√(40^2 +40^2 +2.40.40.(1/2)))      =(√(1600+1600+1600))     =(√(4800 )) N=40(√3) N//
$$\mathrm{F}_{\mathrm{AC}} =\mathrm{k}.\frac{\mathrm{q}_{\mathrm{a}} .\mathrm{q}_{\mathrm{c}} }{\mathrm{r}_{\mathrm{AC}} ^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{9}.\mathrm{10}^{\mathrm{9}} .\frac{\mathrm{2}.\mathrm{10}^{−\mathrm{6}} .\mathrm{2}.\mathrm{10}^{−\mathrm{4}} }{\left(\mathrm{30}.\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{36}}{\mathrm{9}}×\mathrm{10}^{\mathrm{9}−\mathrm{6}−\mathrm{4}+\mathrm{4}−\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{4}×\mathrm{10}^{\mathrm{1}} =\mathrm{40}\:\mathrm{N} \\ $$$$\mathrm{F}_{\mathrm{BC}} =\mathrm{k}.\frac{\mathrm{q}_{\mathrm{b}} .\mathrm{q}_{\mathrm{c}} }{\mathrm{r}_{\mathrm{BC}} ^{\mathrm{2}} }\: \\ $$$$\:\:=\mathrm{9}.\mathrm{10}^{\mathrm{9}} .\frac{\mathrm{2}.\mathrm{10}^{−\mathrm{6}} .\mathrm{2}.\mathrm{10}^{−\mathrm{4}} }{\left(\mathrm{30}.\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{36}}{\mathrm{9}}×\mathrm{10}^{\mathrm{9}−\mathrm{6}−\mathrm{4}+\mathrm{4}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{4}×\mathrm{10}^{\mathrm{1}} =\mathrm{40}\:\mathrm{N} \\ $$$$\mathrm{F}_{\mathrm{R}} =\sqrt{\mathrm{F}_{\mathrm{ac}} ^{\mathrm{2}} +\mathrm{F}_{\mathrm{bc}} ^{\mathrm{2}} +\mathrm{2F}_{\mathrm{ac}} .\mathrm{F}_{\mathrm{bc}} \mathrm{cos}\:\mathrm{60}°} \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{40}^{\mathrm{2}} +\mathrm{40}^{\mathrm{2}} +\mathrm{2}.\mathrm{40}.\mathrm{40}.\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:=\sqrt{\mathrm{1600}+\mathrm{1600}+\mathrm{1600}} \\ $$$$\:\:\:=\sqrt{\mathrm{4800}\:}\:\mathrm{N}=\mathrm{40}\sqrt{\mathrm{3}}\:\mathrm{N}// \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *