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\(\dfrac{109}{6} \div \dfrac{7}{3} = \dfrac{109}{6} \times \dfrac{3}{7} = \dfrac{327}{42} = \dfrac{109}{14} = 7\tfrac{11}{14}\)
Given that $p$ and $q$ are different prime numbers, $r$ is the least prime number greater than $p$, and $s$ is the least prime number greater than $q$.
Compare:
Column A: $r - p$
Column B: $s - q$
The difference between a prime and the next prime varies. For example:
- $3 - 2 = 1$ (gap of 1)
- $5 - 3 = 2$ (gap of 2)
- $7 - 5 = 2$ (gap of 2)
- $11 - 7 = 4$ (gap of 4)
- $23 - 19 = 4$ (gap of 4)
Since $p$ and $q$ are different primes and we don't know which specific primes they are, the gaps $r - p$ and $s - q$ could be equal or different.
For instance:
- If $p = 2$ and $q = 3$: then $r = 3, s = 5$, so $r - p = 1$ and $s - q = 2$, making Column B greater
- If $p = 3$ and $q = 2$: then $r = 5, s = 3$, so $r - p = 2$ and $s - q = 1$, making Column A greater
Since the relationship depends on which primes are chosen, it cannot be determined from the given information.
Using mass ($M$), length ($L$), time ($T$), and electric current ($A$) as fundamental quantities, the dimensions of magnetic permeability are:
Magnetic permeability $\mu_0$ appears in: $F = \dfrac{\mu_0}{4\pi}\dfrac{I_1 I_2 l}{r}$
$[\mu_0] = \dfrac{[F][r]}{[I]^2[l]} = \dfrac{MLT^{-2} \cdot L}{A^2 \cdot L} = MLT^{-2}A^{-2}$
Alternatively, from $B = \mu_0 H$: $[H] = AL^{-1}$, $[B] = MT^{-2}A^{-1}$, so $[\mu_0] = \frac{[B]}{[H]} = \frac{MT^{-2}A^{-1}}{AL^{-1}} = MLT^{-2}A^{-2}$
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