solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
Answer:
2800000J
Explanation:
Parameters given:
Mass = 920kg, weight = 920 * 9.8 = 9016N
Distance = 310m
Angle of inclination = 6.5°
Work done is given as :
W = F*d*cosA
Where A = angle of inclination
W = (9016 * 310 * cos6.5)
W = 2776993.59J
In 2 significant figures, W = 2800000J
Answer:
by a factor of 2
Explanation:
Maximum speed of a body in simple harmonic motion relate to the amplitude by the following formula:
v ( maximum speed in m/s ) = x ( amplitude in meters ) √K /m where K is in N/m and m is kg
v is directly proportional to the amplitude and increases as the amplitude increases by a factor of 2
The correct option is the second one, which is: Every million years.
The impact of a 1-kilometer object on our planet, would cause significant consequences. As the astronomers have found in other terrestrial bodies, this kind of impact creates largers craters on the surface.
Answer:
T1 = 417.48N
T2 = 361.54N
T3 = 208.74N
Explanation:
Using the sin rule to fine the tension in the strings;
Given
amass = 42.6kg
Weight = 42.6 * 9.8 = 417.48N
The third angle will be 180-(60+30)= 90 degrees
Using the sine rule
W/Sin 90 = T3/sin 30 = T2/sin 60
Get T3;
W/Sin 90 = T3/sin 30
417.48/1 = T3/sin30
T3 = 417.48sin30
T3 = 417.48(0.5)
T3 = 208.74N
Also;
W/sin90 = T2/sin 60
417.48/1 = T2/sin60
T2 = 417.48sin60
T2 = 417.48(0.8660)
T2 = 361.54N
The Tension T1 = Weight of the object = 417.48N
