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At the midpoint O (on the equatorial line of both magnets), $r = 0.10\ \text{m}$:
Field due to each magnet on equatorial line: $B = \frac{\mu_0}{4\pi}\frac{M}{r^3}$
$B_1 = \frac{10^{-7} \times 1.20}{(0.10)^3} = 1.20\times10^{-4}\ \text{T}$
$B_2 = \frac{10^{-7} \times 1.00}{(0.10)^3} = 1.00\times10^{-4}\ \text{T}$
Both fields point in the same direction (both N-poles south, equatorial fields add).
Total from magnets $= 2.20\times10^{-4}\ \text{T}$. Adding Earth's component: $2.20 + 0.36 = 2.56\times10^{-4}\ \text{Wb/m}^2$.
A body starts from rest and moves under constant acceleration for 20 s. It covers distance $S_1$ in the first 10 s and $S_2$ in the next 10 s. Then $S_2$ equals:
Using $s = \dfrac{1}{2}at^2$ (from rest):
$S_1 = \dfrac{1}{2}a(10)^2 = 50a$
Total in 20 s: $S_{\text{total}} = \dfrac{1}{2}a(20)^2 = 200a$
$S_2 = 200a - 50a = 150a = 3 \times 50a = \mathbf{3S_1}$
The practice of Flextime is best defined as:
Flextime is the practice of permitting employees to choose, with certain limitations, their own working hours. This helps organizations attract qualified individuals by allowing both employment and family needs to be addressed.
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