Rank the following compounds in order of decreasing vapor pressure.

A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average

Maksim231197 [3]

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

So, knowing $T=619.36 years$ and isolating $a$ from (2) we have:

$a=\sqrt[3]{T^{2}}$ (3)

$a=\sqrt[3]{(619.36 years)^{2}}$ (4)

Finally:

$a=72.66 AU$ T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

$a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km$

4 0

3 years ago

A metal ball rolls from rest at Point A down the track to Point E as shown below.

Tju [1.3M]

Answer:

Explanation:

Velocity is at its greatest when kinetic energy is at its max which is when all the ball's energy has been transformed from potential energy to kinetic energy which is at the lowest point in its travels (assuming the ball is rolling down a ramp). You have no picture here so this answer is a general one, not a specific one.

8 0

2 years ago

A car cruises at a constant rate of 50 miles per hour. How much time will it take to go 600 miles?

Len [333]

Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.

7 0

2 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

- Explain why an aircraft will tend to lose height

olganol [36]

Answer:

the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up

Explanation:

8 0

2 years ago