If 3.5 paper clips = 1.0 pencils and your paper is 1.5 pencils long, how many paper clips long is your paper?

1 answer:

6 0

$\bf \begin{array}{ccll}
paperclips&pencils\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
3.5&1.0\\
p&1.5
\end{array}\implies \cfrac{3.5}{p}=\cfrac{1.0}{1.5}\implies \cfrac{3.5\cdot 1.5}{1}=p$

You might be interested in

A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

lions [1.4K]

Answer:

Approximately $4.61\times 10^{3}\; {\rm N}$ upwards (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ .)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) and
Normal force from the floor (upwards.)

Let $(\text{normal force})$ denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

$\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ .

Let $m$ denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be $(\text{weight}) = m\, g$ .

Let $a$ denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be $(\text{net force}) = m\, a$ .

Rearrange $\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ to obtain an expression for the magnitude of the normal force on this astronaut:

$\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}$ .