A 100 N force pulls a box horizontally across a floor for 2 m. How much was done by the force of gravity (which pulls straight down on the box)?
The work done by the force of gravity will be 200 J.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative. it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force applied = 100 N
d is the displacement = 2 m
W is the work done
The done by the force of gravity will be ;
W = F × d
W= 100 N × 2 m
W = 200 Nm
W = 200 J
Hence, the work done by the force of gravity will be 200 J.
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The price at which a nations currency can be bought using another nations currency is known as
A. The rate of exchange
B.consumer advantage
C.globalization
D.currency trade
Answer:
A. The rate of exchange
Explanation:
The pioneer in the technique of photographic collage was:
The correct answer is Raoul Hausmann. He was an Austrian artist and writer. And he was a forefather and dominant figure of the Dada movement in Berlin, who was known particularly for his ironic photomontages and his provocative writing on art. One of the important figures there also includes the experimental photographic collages, sound poetry and institutional reviews that has a profound influence on the European Avant-Garde in the aftereffects of World War I.
Bright white shadings on infrared images indicate cloud tops that have relatively ________ temperatures
The common balance works on the principle of equality of
Answer:
HAve two arms suspended and one known weight
Explanation:
the common balance is a balance that we all have seen in examples of Justice, it was well known in the old times, but nowadays its more common seeing it as a jsutice symbol, it works in the principle of having two weights accross a balance and letting gravity work on them, while you have an object on one side of the balance of which you know the mass and you add mass to the object in the othar side of the balance to equalize the weights.
PLEASE GO ANSWER MY 3 MOST RECENT QUESTION I NEED HELP!!!!!!!!!!!!!!
3. Compare how a positively charged object and a negatively charged object interact with a neutral object.
4. What happens to the composition of an atom to cause it to become positively or negatively charged?
5. If you rub a balloon on a piece of fabric or carpeting and then hold it against your head, your hair can stand on end. Use what you have learned in this lesson to explain why this happens
You have developed a method in which a paint shaker is used to measure the coefficient of static friction between various objects and a known surface. the shaker oscillates with a fixed amplitude of 40 mm , but you can adjust the frequency of the motion. you have affixed a horizontal tabletop (the known surface) to the shaker so that the tabletop oscillates with it. then you put an object on the tabletop and increase the frequency until the object begins to slip on the surface. part a if a frequency f = 1.75 hz is required before a penny positioned on the tabletop starts to slide, what is the coefficient of static friction between penny and tabletop?
The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.40s
The velocity of the oscillating mass at time [tex]t = 0.40\,{\text{s}}[/tex] is [tex]\boxed{15\times{10^{-2}}\,{\text{m/s}}}[/tex] or [tex]\boxed{15\,{\text{cm/s}}}[/tex] or [tex]\boxed{0.15\,{\text{m/s}}}[/tex].
Further explanation:
Velocity of a particle or a mass at any instant is defined as the rate of change of position of particle with respect to time.
Mathematically,
[tex]V\left( t \right) = \dfrac{d}{{dt}}\left( {X\left( t \right)} \right)[/tex]
If position of a particle or mass is a function of time then velocity of mass at any instant will change with respect to time.
Given:
The position of an oscillating mass varies according to the function [tex]X(t)=({2.0\text{ cm}}})\cos({10t})[/tex].
Mass of an oscillating object is [tex]55\text{ g}[/tex].
Concept:
The velocity of mass at any instant is calculated by using the following relation
[tex]\begin{aligned}V(t)&=\frac{{dX\left( t \right)}}{{dt}}\\&=\frac{d}{{dt}}\left[{\left( {2.0{\kern 1pt} {\text{cm}}} \right)\cos \left( {10t} \right)} \right]\\&=-\left( {20\,{\text{cm/s}}}\right)\sin\left( {10t}\right)\\\end{aligned}[/tex]
Therefore the velocity of the mass at any instant is given by
[tex]V\left( t \right)=-\left( {20{\kern 1pt} {\text{cm/s}}} \right)\sin \left( {10t} \right)[/tex]
From the above expression of velocity it can be observed that velocity is changing with time according to the sin function.
Substitute [tex]0.40\,{\text{s}}[/tex] for t in the above expression
[tex]\begin{aligned}V\left( {0.4\,{\text{s}}} \right)&=-\left( {20\,{\text{cm/s}}} \right)\sin \left( 4 \right)\\&=-\left( {20\,{\text{cm/s}}} \right)\left( { - 0.757} \right)\\&=15.14\,{\text{cm/s}}\\\end{aligned}[/tex]
Thus, the velocity of the oscillating mass at time [tex]t = 0.40\,{\text{s}}[/tex] is [tex]\boxed{15\times{10^{-2}}\,{\text{m/s}}}[/tex] or [tex]\boxed{15\,{\text{cm/s}}}[/tex] or [tex]\boxed{0.15\,{\text{m/s}}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Force
Keywords:
position, oscillating, 55 g, mass, time, x(t)=(2.0cm)cos(10t), t=4 s, determine, velocity, 15 cm/s, 0.15 m/s, rate, change in position.
a speeding car traveling north at 80mph passes the tramway intersection when the police begin pursuit. The police start 6 miles south of tramway traveling at 100mph. How many miles north of tramway will they be when they catch up with the speeding car?
Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 163 v. (b) calculate the speed of an electron that is accelerated through the same potential difference.
Which of these should you not do when merging onto a freeway? A. Select a gap large enough to fit your vehicle. B. Change lanes smoothly. C. Change two lanes at once and speed up. D. Adjust your speed to create a safe following interval.
Efforts to relate gamma-ray bursts with specific sources has had what results so far?
X-rays are high energy electrons that can cause damage when exposed under extreme conditions. The best technology that can block it is using a lightweight type of metal foam. It can take in high energy collisions which also exhibits high forces. it does not only block x-rays but also, neutron radiation and gamma rays. A gamma ray primarily consists of pure energy and no mass is true. In fact it consists of high energy photons and massless. They have no charge and when exposed to any kind of material, it just passes through as if nothing is blocking it.
With the switch closed, how does the voltage across the 20-ω resistor compare to the voltage across the 10-ω resistor?
A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s. How long did the burn phase last?
Battery life is what distinguishes one type of mobile computer from another.
A man stands on a scale in an elevator as shown here. the force of his weight when the elevator is still is fg downward. suppose the elevator's acceleration downward is 1/4 g. the weight of the man, fs, is what
The weight of the man in the elevator is [tex]\frac{3}{4} f_g[/tex].
What is weight?Weight of a body is the force with which the earth attracts it. due to having both magnitude and direction, Weight is a vector quantity. Si unit of weight is Newton.
Given parameter:
The force of his weight in still elevator is = [tex]f_g[/tex]
And, the elevator's acceleration downward is 1/4 g.
Let, the mass of the man is = m.
So, his weight in still elevator = [tex]f_g[/tex] = mg.
Where, g = acceleration due to the gravity.
When the elevator moves downward, the experienced weight of the man will change and magnitude of it will be equal to the difference of his weight in still elevator and pseudo force due to the elevator's acceleration downward is 1/4 g .
The pseudo force due to the elevator's acceleration = mass × acceleration
= m× 1/4 g
= 1/4 mg
Fs=m×g-m×1/4g=m×(g-1/4g)=m×3/4g
Now, the weight of the man experienced during the elevator's acceleration downward is 1/4 g is, [tex]f_s[/tex] = [tex]f_g -\frac{1}{4}[/tex]mg = 3/4 mg = [tex]\frac{3}{4}[/tex] [tex]f_g[/tex].
Hence, the net weight of the man is [tex]\frac{3}{4}[/tex] [tex]f_g[/tex].
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What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 35.0 cm?
To calculate the force exerted by the water on the window of an underwater vehicle at a certain depth, use the formula Force = Pressure x Area. Calculate the pressure using the formula Pressure = Density x Gravity x Depth. Calculate the area using the formula Area = π x (radius)^2.
Explanation:The force exerted by the water on the window of an underwater vehicle at a certain depth can be calculated using the formula:
Force = Pressure x Area
The pressure exerted by water at a given depth is given by the formula:
Pressure = Density x Gravity x Depth
Using the given diameter of the window, you can calculate the area of the window by using the formula for the area of a circle:
Area = π x (radius)^2
By substituting the values into the formulas and calculating, you can find the force exerted by the water on the window.
Harry is reading an online summary of the law of reflection. The site states that after light hits a mirror, the angle of reflection is the angle between the reflected ray and the normal, which is the surface of the mirror.Which statement corrects an error on the site?
'A line perpendicular to the mirror's surface is called the normal. This is the right statement for correcting a website error.
What is the law of reflection?The law of reflection states that when a ray is reflected off a downy surface, the reflected ray's slope is equivalent to the incident ray's slope.
The incident ray and the reflected ray are always in the same plane as the incident ray and perpendicular to the surface at the incident ray's point of reference.
The angle of reflection is comparable to the angle of incidence when light rays fall on a flat surface, and the incident ray, reflected ray, and normal to the surface all lie in the same plane.
Hence 'A line perpendicular to the mirror's surface is called the normal. This is the right statement for correcting a website error.
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A bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high. if air resistance is negligible, what is the horizontal distance the bullet travels before hitting the ground?
Final answer:
To find the horizontal distance a bullet travels before hitting the ground when fired from a certain height, one must calculate the time of fall due to gravity and then multiply by the horizontal velocity. In the given scenario, the bullet would travel approximately 216.5 meters.
Explanation:
The problem of calculating the horizontal distance a bullet travels before hitting the ground involves understanding projectile motion, specifically when an object is fired horizontally from a certain height. The question states that a bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high, and we need to assume air resistance is negligible. To find the horizontal distance, we first need to determine the time it takes for the bullet to reach the ground.
In the absence of air resistance, the horizontal motion of the bullet is at a constant velocity. Meanwhile, the vertical motion is influenced by gravity and can be calculated using the following kinematic equation for free fall:
t =√(2h/g)
Where:
t is the time in seconds
h is the height in meters
g is the acceleration due to gravity (approximately 9.81 m/s²)
Once we have the time, we can calculate the horizontal distance using the equation:
x = v * t
Where:
x is the horizontal distance
v is the horizontal velocity
t is the time
Applying these calculations:
t = √(2 * 11 m / 9.81 m/s²) = √(2.24 s²) = 1.497 s (approximately)
Then:
x = 144.7 m/s * 1.497 s = 216.5329 m (approximately)
Therefore, the bullet will travel approximately 216.5 meters horizontally before hitting the ground.
Which statement best describes the difference between acceleration and deceleration?
Deceleration is a type of acceleration where an object slows down, always acting opposite to the direction of motion. Negative acceleration, on the other hand, refers to acceleration in a negative direction as per the coordinate system and may not involve slowing down. Correctly distinguishing between these terms requires understanding both the motion's direction and the chosen coordinate system.
Explanation:The difference between acceleration and deceleration relates to the direction of the change in velocity relative to the object's current motion. While acceleration refers to a change in velocity, either in magnitude or direction, deceleration specifically refers to acceleration that occurs in the opposite direction of an object's motion and results in a reduction of the object's speed. It is a common misconception to equate deceleration with negative acceleration; although related, they are not always the same because negative acceleration refers to acceleration in a negative direction in the chosen coordinate system, which might not necessarily mean the object is slowing down.
Deceleration always reduces speed, whereas negative acceleration can occur when an object is speeding up in a direction defined as negative by a coordinate system. Therefore, deceleration is a form of acceleration that always acts in the opposite direction to the velocity, causing the object to slow down. On the other hand, negative acceleration is coordinate-dependent and may either represent an object slowing down or speeding up, depending on its initial direction of motion.
It is advisable to avoid the use of the term 'deceleration' when referring to negative acceleration, as one must also consider the sign of the object's velocity to determine if the object is slowing down.
A metal smith pours 3.00 kg of lead shot at 99oc into 1.00 kg of water at 25oc in an isolated container. what is the final temperature of the mixture?
Suppose you are an astronaut and you have been stationed on a distant planet. you would like to find the acceleration due to the gravitational force on this planet so you devise an experiment. you throw a rock up in the air with an initial velocity of 11 m/s and use a stopwatch to record the time it takes to hit the ground. if it takes 7.0 s for the rock to return to the same location from which it was released, what is the acceleration due to gravity on the planet?
Answer: g = -3.143 m/s²
Explanation: To determine acceleration due to the gravitational force, it can be used the following formula:
v = v₀ + gt, where:
v₀ is the initial velocity;
g is acceleration due to gravity
t is the time to return to the point of origin
When the rock return to the point of origin, there is no velocity, so v = 0.
The time to go up is the same to go down, so t = [tex]\frac{t}{2}[/tex] = [tex]\frac{7}{2}[/tex].
Substituing in the formula:
v = v₀ + gt
0 = 11 + g.[tex]\frac{7}{2}[/tex]
g = [tex]\frac{(0 - 11).2}{7}[/tex]
g = - 3.143
The acceleration due to the gravitational force is g = - 3.143 m/s².
Astronaut mark uri is space-traveling from planet x to planet y at a speed of relative to the planets, which are at rest relative to each other. when he is precisely halfway between the planets, a distance of 1.0 light-hour from each one as measured in the planet frame, nuclear devices are detonated on both planets. the explosions are simultaneous in mark's frame. what is the difference in the time of arrival of the flashes from the explosions as observed by mark?
The time of arrival of the flashes from the explosions will be different for Mark due to time dilation. Mark's velocity relative to the planets will cause a time difference between the observed arrival times. The Lorentz transformation formula can be used to calculate the time dilation factor.
Explanation:The difference in the time of arrival of the flashes from the explosions as observed by Mark can be calculated using the concept of time dilation. According to the theory of special relativity, time is relative and depends on the observer's frame of reference. As Mark is traveling at a speed relativistic to the planets, the time measured by Mark will be different from the time measured by an observer at rest on the planets.
In this scenario, Mark is traveling halfway between the planets, so the distance to each planet from Mark is 1.0 light-hour. The explosions on both planets are simultaneous according to Mark's frame of reference. However, due to time dilation, the time of arrival of the flashes from the explosions will be different as observed by Mark.
The time dilation factor can be calculated using the Lorentz transformation formula:
t' = t * sqrt(1 - (v^2 / c^2))
Where:
t' is the time measured by Mark
t is the time measured in the planet frame
v is Mark's velocity relative to the planets
c is the speed of light
Since Mark is traveling at a speed relativistic to the planets, his velocity v will be a significant fraction of the speed of light, resulting in a noticeable time dilation effect.
After an average acceleration of 3.95 m/s2 during 2.25 s, a car reaches a velocity of 15.3 m/s. find the car\'s initial velocity.
The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. to the crew’s great sur- prise, a klingon ship is 100 km directly ahead, traveling in the same direction at a mere 20 km/s. without evasive action, the enterprise will overtake and collide with the klingons in just slightly over 3.0 s. the enterprise’s computers react instantly to brake the ship. what magnitude acceleration does the enterprise need to just barely avoid a collision with the klingon ship? assume the acceleration is constant.
The magnitude of the acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship is 3.27 x 10^10 m/s^2.
Explanation:To find the magnitude acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship, we can use the impulse-momentum relation. The change in momentum can be determined by subtracting the initial momentum from the final momentum. The mass of the Enterprise is given as 2 x 10^9 kg. The initial momentum can be calculated by multiplying the mass of the Enterprise by its initial speed, and the final momentum can be calculated by multiplying the mass of the Klingon ship by its final speed. Using the formula for impulse, which is the change in momentum divided by the time interval, we can solve for the acceleration.
The initial momentum of the Enterprise is (2 x 10^9 kg) x (50 km/s) = 1 x 10^11 kg m/s. The final momentum of the Klingon ship is (100 km) x (20 km/s) = 2 x 10^8 kg m/s. The change in momentum is (2 x 10^8 kg m/s) - (1 x 10^11 kg m/s) = -9.80 x 10^10 kg m/s. The time interval is given as 3.0 s.
Using the formula for impulse, we have:
Force x Time Interval = Change in Momentum
Force = Change in Momentum / Time Interval
Force = (-9.80 x 10^10 kg m/s) / (3.0 s) = -3.27 x 10^10 N
The magnitude of the acceleration the Enterprise needs to just barely avoid a collision with the Klingon ship is 3.27 x 10^10 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity of the Enterprise.
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To avoid the collision, the Enterprise needs an acceleration of approximately 2.22 km/s². This calculation involves initial speeds, distances, and applying the constant acceleration formula. The relative speed and distance are critical factors in determining the necessary braking acceleration.
To prevent a collision with the Klingon ship, we need to determine the magnitude of acceleration required by the starship Enterprise. Here’s a step-by-step breakdown:
The initial relative speed between the Enterprise and the Klingon ship: [tex]\Delta v = 50 km/s - 20 km/s = 30 km/s.[/tex]The distance to the Klingon ship:[tex]d = 100 km[/tex].The time to collision without evasive action:[tex]t = 3 s[/tex].Using the equation for constant acceleration: [tex]d = v_i * t + 0.5 * a * t^2[/tex]. Here, d is the distance, v_i is the initial speed, a is acceleration, and t is time.Since the Enterprise needs to stop before reaching 100 km, [tex]v_i * t[/tex] must be subtracted from d:[tex]100 km = (30 km/s) * 3 s + 0.5 * a * (3 s)^2[/tex]
Simplifying the equation:
[tex]100 km = 90 km + 4.5a[/tex]
Solve for a:
[tex]10 km = 4.5a[/tex]
[tex]a = - (10 km) / (4.5 s^2) = -2.22 km/s^2[/tex]
Thus, the magnitude of the required acceleration is approximately 2.22 km/s².
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:
density (salt water) = 1029 kg / m^3
density (fresh water) = 1000 kg / m^3
The amount of water (in mass) displaced by the barge should be equal in two waters.
mass displaced (salt water) = mass displaced (fresh water)
Since mass is also the product of density and volume, therefore:
[density * volume]_salt water = [density * volume]_fresh water ---> 1
First we calculate the amount of volume displaced in the harbour (salt water):
V = 3.0 m * 20.0 m * 0.70 m
V = 42 m^3 of salt water
Plugging in the values into equation 1:
1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water
Volume fresh water displaced = 43.218 m^3
Therefore the depth of the barge in the river is:
43.218 m^3 = 3.0 m * 20.0 m * h
h = 0.72 m (ANSWER)
Based on Archimedes' principle, the rectangular barge will sit at the same depth of 0.70m in both the harbor and the river, if the river and the harbor have the same type of water.
Explanation:The depth at which the rectangular barge will sit in the river relates to the concept of buoyancy and the principle of Archimedes. According to this principle, the weight of the water displaced by the barge is equal to the weight of the barge itself.
Provided that the river and the harbor have the same type of water (salt water or fresh water), the barge will sit at the same depth in both, which is 0.70m deep. This is because the barge, with dimensions of 3.0m by 20.0m, will displace an equal volume of water to balance its own weight in any body of water it is placed in.
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Starting from rest, a disk takes 8 revolutions to reach an angular velocity ω at constant angular acceleration. how many additional revolutions are required to reach an angular velocity of 3 ω ?
To find the additional revolutions required to reach an angular velocity of 3ω, we can use the equations of rotational motion.
Explanation:To find the solution to this problem, we can use the equations of rotational motion. The first step is to find the initial angular velocity, ω0. Given that the disk takes 8 revolutions to reach an angular velocity ω, we can calculate that ω0 = 2π * 8 / 8 = 2π rad/s.
Next, we can use the equation ω = ω0 + αt to find the angular acceleration, α. Rearranging the equation, we get α = (ω - ω0) / t = (3ω - 2π) / t.
Finally, we can use the equation ω = ω0 + αt to find the additional time, t', required to reach an angular velocity of 3ω. Rearranging the equation, we get t' = (3ω - ω0) / α = (3ω - 2π) / ((3ω - 2π) / t).
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A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r with arrowi = (12.0 i − 2.00 j) m. after the fish swims with constant acceleration for 17.0 s, its velocity is v with arrow = (23.0 i − 1.00 j) m/s. (a) what are the components of the acceleration of the fish? ax = m/s2
Final answer:
To find the acceleration components of the fish, the change in velocity components is divided by the time interval. The x-component is 1.12 m/s² and the y-component is -0.12 m/s².
Explanation:
The student is asking how to calculate the components of acceleration given the initial and final velocity of a fish over a certain time period. To find the acceleration components, we use the formula a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time over which the change occurred. For the x-component: ax = (23.0 m/s - 4.00 m/s) / 17.0 s = 1.12 m/s². For the y-component: ay = (-1.00 m/s - 1.00 m/s) / 17.0 s = -0.12 m/s².
Two children standing on opposite sides of a merry-go-round are trying to rotate it. they each push in opposite directions with forces of magnitude 10.2 n. (a) if the merry-go-round has a mass of 180 kg and a radius of 1.8 m, what is the angular acceleration of the merry-go-round? (assume the merry-go-round is a uniform disk.)
The angular acceleration of the merry-go-round can be found using Newton's second law for rotation and the moment of inertia. To calculate the total moment of inertia, we can approximate the child as a point mass and add it to the moment of inertia of the merry-go-round which is 1.8
Explanation:The angular acceleration of the merry-go-round can be found using Newton's second law for rotation which states that the torque is equal to the moment of inertia times the angular acceleration. In this case, the torque is equal to the product of the force applied by each child and the radius of the merry-go-round. The moment of inertia of the merry-go-round can be found using the equation [tex]I = 1/2 * MR^2,[/tex] where M is the mass and R is the radius of the merry-go-round.
Thus,
F = m a
a = F / m
a = 10.2 N / 180 kg
[tex]a = 0.057 m / s^2[/tex]
The relationship between angular velocity (a) and angular velocity (ω) is:
w = a / r
[tex]w = (0.057 m / s^2) / 1.8 m[/tex]
[tex]w = 0.031 rad / s^2[/tex]
To get an answer in terms of degrees per s^2, we multiply it with the conversion factor = 180˚ / π
[tex]w = (0.031 rad / s^2) (180 / \pi rad)[/tex]
[tex]w = 1.8 / s^2[/tex]
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At a certain temperature, a 29.5-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 22.6 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients)?
The equilibrium constant for the reaction of SO2 with N2O to form SO3 and N2 is approximately 22.2.
The given problem involves finding the equilibrium constant for the reaction of SO2 with N2O to form SO3 and N2. The balanced chemical equation is:
SO2 (g) + N2O (g) ↔ SO3 (g) + N2 (g)
First, we need to calculate the number of moles of each gas using their masses and molar masses:
[tex]\text{Molar mass of }SO_2: 64.07 g/mol\\\text{Molar mass of }SO3: 80.07 g/mol\\\text{Molar mass of }N_2: 28.02 g/mol\\\text{Molar mass of }N_2O: 44.01 g/mol[/tex]
Using the formula: moles = mass / molar mass, we find:
[tex]\text{Moles of }SO_2 = 4.6 g / 64.07 g/mol = 0.0718 mol\\\text{Moles of }SO_3 = 3.5 g / 80.07 g/mol = 0.0437 mol\\\text{Moles of }N_2 = 22.6 g / 28.02 g/mol = 0.8065 mol\\\text{Moles of }N_2O = 0.98 g / 44.01 g/mol = 0.0223 mol[/tex]
The equilibrium constant expression, [tex]K_c[/tex], for the reaction is:
[tex]K_c = [SO_3][N_2] / [SO_2][N_2O][/tex]
In a 29.5 L container, the concentrations of each gas are:
[tex]SO_2= 0.0718 mol / 29.5 L = 0.00243 M\\SO_3 = 0.0437 mol / 29.5 L = 0.00148 M\\N_2 = 0.8065 mol / 29.5 L = 0.0273 M\\N_2O= 0.0223 mol / 29.5 L = 0.000756 M[/tex]
Now substitute these concentrations into the equilibrium expression:
[tex]Kc = (0.00148 M) \times (0.0273 M) / (0.00243 M) \times (0.000756 M)\\Kc = 22.2[/tex]