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4 years ago

What are the factors affect the vapor pressure of a liquid?

Physics

1 answer:

hodyreva [135]4 years ago

4 0

Answer:

surface area

inter molecular forces

temperature

Explanation:

vapor pressure is the pressure caused by the evaporation of liquids and is affected by the following:

surface area

inter molecular forces

temperature

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A car traveling at 27 m/s runs out of gas while traveling up a 8.0 ∘ slope. How far will it coast before starting to roll back d

Ad libitum [116K]

Answer:

Approximately $2.7 \times 10^{2}\; \rm m$ along the slope, assuming that no energy was lost to friction.

Explanation:

Let $v$ denote the initial velocity of this vehicle. Let $m$ denote the mass of this vehicle. The kinetic energy (KE) of this vehicle would have initially been:

$\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

The gain in the gravitational potential energy (GPE) of this vehicle is proportional to the increase in its height.

Let $g$ denote the gravitational field strength. $g \approx 9.81\; \rm m \cdot s^{2}$ on the earth. If the increase in the height of this vehicle is $h$ , this vehicle would have gained GPE:

$m \cdot g \cdot h$ .

Hence, the height of this vehicle is maximal when the GPE of this vehicle is maximized.

Since the vehicle went out of fuel, all its GPE would have been converted from KE. Assuming that no energy was converted to friction. The GPE of this vehicle would be maximal when the entirety of the KE was converted to GPE.

Hence, the maximal GPE of this vehicle would be equal to its initial KE:

$\displaystyle m \cdot g \cdot h = \frac{1}{2}\, m \cdot v^{2}$ .

The maximum height of this vehicle would be:

$\begin{aligned} h &= \frac{(1/2) \, m \cdot v^{2}}{m \cdot g}\\ &= \frac{v^{2}}{2\, g}\end{aligned}$ .

Given that $v = 27\; \rm m\cdot s^{-1}$ , the maximum height of this vehicle would be:

$\begin{aligned} & \frac{v^{2}}{2\, g} \\ =\; & \frac{(27\; \rm m\cdot s^{-1})^{2}}{2 \times 9.81\; \rm m\cdot s^{-2}} \\ \approx \; &37.2\; \rm m\cdot s^{-1}\end{aligned}$ .

Refer to the diagram attached. The distance that this vehicle traveled along the slope would be approximately:

$\begin{aligned}\frac{37.2\; \rm m}{\sin(8.0^{\circ})} \approx 2.7 \times 10^{2}\; \rm m\end{aligned}$ .

6 0

3 years ago

Particles q1, 92, and q3 are in a straight line.

Kisachek [45]

The answer to the question is 1.125 Newton

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.

Electrostatic force between q1 and q3

r will be 0.500 + 0.500 = 1 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²

F₁₃ = 2.25 × 10⁻¹ N

Electrostatic force between q2 and q3

r will be 0.500 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²

F = (225 × 10⁻³) / (25 × 10⁻²)

F₂₃ = 9 × 10⁻¹ N

Total force on q3 will be F₁₃ + F₂₃

Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 = ( 11.25 × 10⁻¹ ) N

Total force on q3 = 1.125 N

Thus after solving we got the net force on q3 as 1.125 Newton

Learn more about Electrostatic Force here:

brainly.com/question/23121845

#SPJ10

7 0

2 years ago

A 10-cm-diameter circular loop of wire is placed in a 0.84-T magnetic field. Part A When the plane of the loop is perpendicular

Oksanka [162]

Answer:

When the plane of the loop is perpendicular to the field lines, the magnetic flux through the loop is 6.284 x 10⁻³ wb.

Explanation:

The magnetic flux through the loop is the product of magnetic field strength, circular area of the loop and angle of inclination.

Ф = BAcosθ

When the plane of the loop is perpendicular to the field lines, θ = 0

Ф = BAcos0

Ф = BA

Where;

Ф is the magnetic flux through the loop

B is the magnetic field strength = 0.8 T

A is the circular area of the loop;

$A =\frac{\pi D^2}{4} = \frac{\pi *0.1^2}{4} = 0.007855 m^2$

Ф = 0.8 x 0.007855 = 6.284 x 10⁻³ wb

Therefore, when the plane of the loop is perpendicular to the field lines, the magnetic flux through the loop is 6.284 x 10⁻³ wb.

5 0

3 years ago

Read 2 more answers

Urgent! ASAP!! Please provide explanation so I can also learn it

pantera1 [17]

Answer:

3N not 0N (first blank)

3N (second blank)

0N (third blank)

Explanation:

1st blank - If a force equals 0, it is balanced.

2nd blank - If one force of 3 Newtons is pushing to the left and another force of 3 Newtons is pushing from the right it wouldn't go anywhere, so it would be balanced at zero.

3rd blank - It would be 0N because forces of the same amount pushing opposite directions would be balanced.

7 0

4 years ago

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

3 years ago