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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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Katrina’s recipe for orange muffins calls for 400 grams of flour. How many kilograms of flour are in the recipe?

aliina [53]

Answer:

0.4 Kilograms of flour

Explanation:

If you convert 400 grams into kilograms (400 ÷ 1,000), you would get 0.4 kilograms.

3 0

2 years ago

A student writes the following script for a scene in a futuristic movie.

Triss [41]

The astronauts hearing the rocket landing is inaccurate. Sound waves can’t travel through a vacuum, meaning a place where there are no air particles. They are mechanical waves and require a medium to travel through. So, because there is no air in space, you can’t hear anything.

8 0

2 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

3 years ago

Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field

ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C

Explanation:

a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

r² = x² + y²

E1 = k q / (x² + y²)

Particle 2. located at x = -24.9 m

r² = x² + y²

E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

cos θ = CA / H = x / (x2 + y2) ½

E1x = E1 cos θ

E2x = = E1 cos θ

The resulting field

Ey = 0

Ex = E1x + E2x 2 E1x

Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

Ex = 497.15 x / (x² + 0.062) ³/₂

Point where can the value of the electric field x = 38.1 cm = 0.381 m

Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂

Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

Ex= 189.41 /0.0943

Ex = 2008 N / C

c) E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

Ex = 497.15 x / (x² + 0.062) ³/₂

Let's use some arithmetic

Ex / 497.15 = x / (x² + 0.062) ³/₂

[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)

The roots of this equation are the solution to the problem,

For Ex = 1.00 kN / C = 1000 N / C

[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

1.312 = (∛x² ) / (x² + 0.062)

1.312 (x² + 0.062) = ∛x²

1.312 X² - ∛x² + 1.312 0.062 = 0

1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0

2 years ago

Why is the use of carbon-14 limited

Luden [163]

Carbon-14 is naturally created with the interaction of high-energy cosmic rays with atmospheric nitrogen. As part of the atmosphere, living organisms take in the carbon and incorporate this into living tissues. As long as the organism is alive and breathing, it keeps adding new carbon-14. When the organism dies, it stops gaining carbon-14 - or anything else, of course.

Carbon-14 is slightly radioactive, with a half-life of about 5700 years. If we assume that the atmospheric production of carbon-14 has been steady for the last 100,000 years, we can calculate the approximate age of when the organism died by determining what percentage of carbon-14 still exists in the dead material.

Paleo-archaeologists and anthropologists use this information when studying old cultures and civilizations.

8 0

2 years ago