1) science does not accept personal story's as evidence, pseudoscience relies on these story's as evidence.
2) science argues from scientific knowledge, pseudoscience argues from ignorance
3) science progresses, pseudoscience does not progress
and 4) (just in case) science holds pier review, pseudoscience does not

8 0

4 years ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

2 years ago

Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm

Mamont248 [21]

Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8

3 0

3 years ago