<span>1.В,2.Б,3.В, </span>
<span>4 дано:</span>
І1=5А
<span>І2=10А</span>
<span>t</span>=0,1<span>c</span>
<em><span>ε</span></em><span /><span>=20</span><span>B</span>
<span> L</span><span>-?</span>
<em><span>ε</span></em><span /><span>=- </span><span>L</span><em><span><span> </span>∆ </span></em><em><span>I</span></em><em><span>∆</span></em><em><span>t</span></em><span>, </span><span><span> </span>L</span><span>=</span><em><span><span>ε</span></span></em><span>*</span><em><span>∆</span></em><span>t/</span><em><span>∆</span></em><span>I</span>
<span>L=20*0,1/10=0,2Гн</span>
<span>5дано:</span>
<span>L=3Гн</span>
<em><span>ε</span></em><span /><span>=</span><span>15</span><span>B</span>
<span>I=50A</span>
<em><span>∆</span></em><span>t-?</span>
<em><span>ε</span></em><span /><span>=- </span><span>L</span><em><span><span> </span>∆ </span></em><em><span>I</span></em><em><span>∆</span></em><em><span>t</span></em><span /><span>, </span><span><span> </span></span><em><span>∆</span></em><span>t=L*</span><em><span>∆</span></em><span>I/</span><em><span> ε</span></em><span /><span>, </span><em><span>∆</span></em><span>t=3*50/15=10c</span>
<span> </span>
силу находим по закону Гука F=kx=250 Н/м*0.03 м=
k-коэфициент жесткости пружины равен 250 Н/м
x-растяжение пружины на 3 см=0.03 м
ускорение находим по 2 -му закону Ньютона а=F/m=250 Н/м*0.03 м/5 кг=1.5 м/с^2
M=d*F
F=M/d
F=2 Н ..............................
