Question

show all work for full credit.

1. solve for the indicated variable: 2 marks each

a) $ma = f-mu mg$; $m$
b) $f = \frac{gm_1m_2}{r^2}$; $m_1$

1. convert the following measurements: 2 marks each

a) $96.5\\ mhz=$______$mhz$
b) $289\\ mu m/s=$______$km/min$
c) the physicist john von neumann once argued that no lecture should be longer than a microcentury. how long is one of our 80 - minute classes in microcenturies?

Explanation:

1. a) Solve \(ma = F-\mu mg\) for \(m\)

Step1: 移項

將含\(m\)的項移到一邊，得到\(ma+\mu mg = F\)。

Step2: 提取公因式

提取公因式\(m\)，即\(m(a + \mu g)=F\)。

Step3: 求解\(m\)

兩邊同時除以\((a+\mu g)\)，可得\(m=\frac{F}{a + \mu g}\)。

1. b) Solve \(F=\frac{Gm_1m_2}{r^2}\) for \(m_1\)

Step1: 等式兩邊同乘\(r^2\)

得到\(Fr^2=Gm_1m_2\)。

Step2: 求解\(m_1\)

兩邊同時除以\(Gm_2\)，即\(m_1=\frac{Fr^2}{Gm_2}\)。

2. a) Convert \(96.5\ MHz\) to \(mHz\)

Step1: 明確單位換算關係

\(1\ MHz = 10^{6}\ Hz\)，\(1\ mHz=10^{- 3}\ Hz\)，所以\(1\ MHz = 10^{9}\ mHz\)。

Step2: 計算

\(96.5\ MHz=96.5\times10^{9}\ mHz = 9.65\times 10^{10}\ mHz\)。

2. b) Convert \(289\ \mu m/s\) to \(km/min\)

Answer:

1. a) \(m=\frac{F}{a + \mu g}\)
2. b) \(m_1=\frac{Fr^2}{Gm_2}\)
3. a) \(9.65\times 10^{10}\ mHz\)
4. b) \(1.734\times10^{-8}\ km/min\)
5. c) Approximately \(1.52\) micro - centuries