The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
the number of significant figures in the measurement 4.300×10^5 km are
Answer:
6
Explanation
Any numbers in scientific notation are considered significant. For example, 4.300 x 10-4 has 4 significant figures.
Answer From Gauth Math
Solar System - Scaling. When you look at Neptune in a telescope, you are actually looking into the past as the light has to travel from Neptune to your eyes. If the speed of light is ~300,000 km/s, how far back into the past are you looking (or put another way, how long does it take light to travel from Neptune to your eyes on Earth)
Answer:
Distance from sun to Neptune = 4.495E9 km
Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec
That is from sun to Neptune time fof light = 250 min
Time for light to travel from sun to earth is about 8 min
So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and
250 min / 8 min is roughly 30
The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:
t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s
depending on the relative distance of the two planets
given parameters
The speed of light c = 300,000 km / s = 3 10⁸ m / s The distance from Neptune to Sum
to find
The time when light arrives from Neptune to Earth
They velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships
v = d / t
t = d / v
where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)
We look in the tables for the distances and the rotation periods around the sun
distance ( m) period (s)
Sun Neptunium 4.50 10¹² 5.2 10⁹
Sun - Earth 1.5 10¹¹ 3.2 10⁷
With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:
Maximum approach
positions relative distance from the dos Plantetas is
Δd = [tex]x_{Neptuno - Sum} - x_{Earth - Sum}[/tex]d
Δd = 4.50 10¹² - 1.5 10¹¹
Δd = 43.5 10¹¹ m
the time it takes for Neptune's light to reach Earth is
Δt = [tex]\frac{ 43.5 \ 10^{11} }{3 \ 10^8}[/tex]
Δt = 14.5 10³ s
Δt = 1.45 10⁴ s
We reduce to hours
Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h
Maximum away
Δd = [tex]x_{Neptune - Sum} + x_{Neptune-Sum}[/tex]
Δd = 4.50 10¹² + 1.5 10¹¹
Δd = 46.5 10¹¹
The time is
Δt = [tex]\frac{46.5 \ 10^{11}}{ 3 \ 10^8}[/tex]
Δt = 15.5 10³
Δt = 1.55 10⁴ s
We reduce to hours
Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h
In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:
t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s
depending on the relative distance of the two plants
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No matter how far you stands from a mirror your image appear errect .the mirror is
Answer:
convex mirror
.....................
Answer:
convex mirror..........
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.
The velocity of the ball is 100m/s
The first step is to write out the parameters;
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
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a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.
Answer:
Florida is tropical, with a significant rainy seson
In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?
a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time
Answer:
b. The heavier one reaches the bottom first.
Answer:
B
Explanation:
The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
What is an internal resistance?
Explanation:
some thing inside a resistor
PLEASE ANSWER IF YOU CAN AND NOT FOR THE SAKE OF GAINING POINTS!
Determine the point of contraflexure
Answer:
The point of contraflexure (PoC) occurs where bending is zero and at the point of change between positive and negative (or between compression and tension). In a beam that is flexing (or bending), the point where there is zero bending moment is called the point of contraflexure.
define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?
When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
You are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.
The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.
Answer:
The answer is "0.00466 N".
Explanation:
[tex]F=(B \times i) L\\\\[/tex]
therefore the smaller side is parallel to magnetic field
[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]
calculating the force on the layer side:
[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]
Therefore [tex]F_o[/tex] the net force on the rectangular loop [tex]= 0.00466 \ N[/tex]
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
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cyclist always bends when moving the direction opposite to the wind. Give reasons
Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2
The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶ W/m².
The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.
a) Intensity of the electromagnetic wave for the electromagnetic field.
The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:
[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex] (1)
Where:
c: is the speed of light = 3.00*10⁸ m/s
E: is the magnitude of the electric field = 1.5 V/m
ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²
Hence, the intensity of the electromagnetic wave (eq 1) is:
[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]
b) Intensity of the electromagnetic wave for the magnetic field
We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:
[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex] (2)
Where:
B: is the magnitude of the magnetic field = 10⁻¹⁰ T
μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A
Therefore, the intensity of the electromagnetic wave (eq 2) is:
[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]
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I hope it helps you!
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
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I only need help with e (bottom of the page).
Explanation:
The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by
[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]
[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]
[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]
The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by
[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]
[tex]\:\:\:\:= 26.1°[/tex]
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.
I will give brainliest!
P.S. AL2006 if you see this please help!
I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.
"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .
I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.
Answer:
1 -> 2 & 3 -> 4: Isobaric process
4 -> 1: Work done BY the system
2 -> 3: Work done ON the system
W(total): W = 0J
Explanation:
The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.
I hope this helps :))
Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answer:
PlROCA
Explanation:
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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The timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500 micro F capacitor and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?
Answer:
check 2 photos for answer
check 2 photos for answer
Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.
Answer:
T1 = 736.6 N, T2 = 193.5 N
Explanation:
W = 76 N
The tension is T1 and T2.
By use of Lami's theorem
[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]
A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.
The horizontal force applied to the lower block is approximately 1,420.85 Newtons
The known parameters are;
The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N
The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N
The length of the string attached to the block, W₂, l = 6 m
The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m
The coefficient of friction between the surfaces, μ = 0.25
Let T represent the tension in the string
The upward force on W₂ due to the string = T × sin(θ)
The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)
The tension in the string, T = N₂ × μ × cos(θ)
∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)
sin(θ) = √(6² - 5²)/6
cos(θ) = 5/6
∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
T ≈ 183.27 N
The normal reaction on W₂, N₂ = T/(μ × cos(θ))
∴ N₂ = 183.27/(0.25 × 5/6) = 879.7
N₂ ≈ 879.7 N
The friction force, [tex]F_{f2}[/tex] = N₂ × μ
∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N
The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂
[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N
The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ
∴ [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N
The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = 1,200.925 N + 219.925 N = 1,420.85 N
The horizontal force applied to the lower block, P ≈ 1,420.85 N
