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3 years ago

Write the equality and conversion factors for the relationship between miles and hours for a car traveling at 55 mi/h

Physics

1 answer:

AysviL [449]3 years ago

5 0

Using Raoult’s Law, we can calculate for the vapor pressure using the formula:

P = x * P(pure)

Where x is the mole fraction of water, P(pure) = pressure of pure water, P = pressure when in solution

calculating for mole fraction of water:

x = (0.42kg / 18kg/kmol) / ((0.42kg / 18kg/kmol) + (0.075kg / 192.12kg/kmol))

x = 0.9835

Therefore:

P = 0.9835* P(71.93 mmHg)

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A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

3 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

An increase in resistance causes the flow of electric current to

Zolol [24]

An increase in voltage causes the flow of electric current to INCREASE. An increase in resistance causes the flow of electric current to DECREASE.

100% right

6 0

3 years ago

Read 2 more answers

A light source emits light with dominant wavelengths in the range of 650 to 690 nm. what is the principle color of light emitted

dsp73

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The visible range extends roughly from 400 nm (violet) to 700 nm (red).
Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
The wavelengths in the range of 650 to 690 nm have red as the dominant color.

6 0

3 years ago

Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and c

rodikova [14]

Answer:

Pressure of woman will be $99.87\times 10^5N/m^2$