The temperature outside is 1.6 degrees Celsius
What is the Charles law?Charles's law, also known as the law of volumes, is a fundamental gas law that describes the relationship between the volume and temperature of a gas at a constant pressure.
It states that, for a fixed mass of gas at a constant pressure, the volume of the gas is directly proportional to its absolute temperature.
By the use of the Charlee's law;
V1/T1 = V2/T2
V1T2 = V2T1
T2 = V2T1/V1
T2 = 4.9 * 297/5.3
T2 = 1.6 degrees Celsius
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What two structures help ecoli move
Escherichia coli use long, whip-like structures called flagella to propel themselves.
What is E.coli.?Escherichia coli, also known as E. coli, is a Gram-negative, facultative anaerobic, rod-shaped, coliform bacterium of the genus Escherichia that is commonly found in the lower intestine of warm-blooded organisms
Given is to find what two structures help E.coli move.
Escherichia coli use long, whip-like structures called flagella to propel themselves. Motors in the cell's wall spin the flagella into bundles that rotate counter-clockwise, creating a twist that causes the bacterium to rotate clockwise, or towards the right when viewed from above.
Therefore, Escherichia coli use long, whip-like structures called flagella to propel themselves.
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Calculate the resultant of the pair of velocities 120 km/hkm/h north and 72 km/hkm/h west.
Express your answer to two significant figures and include the appropriate units.
Calculate the resultant if both of the velocities are directed north.
Express your answer to two significant figures and include the appropriate units.
The resultant velocity of two significant figures is 192 km/h.
To calculate the resultant of the pair of velocities, we need to use the Pythagorean theorem.
The Pythagorean theorem states that for any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the hypotenuse is the resultant velocity, and the other two sides are the given velocities.
So, the resultant velocity is given by:
Resultant velocity = [tex]\sqrt{((120 km/h)^2 + (72 km/h)^2)[/tex]
Resultant velocity = [tex]\sqrt{(14400 km^2/h^2 + 5184 km^2/h^2)[/tex]
Resultant velocity = [tex]\sqrt{(19584 km^2/h^2)[/tex]
Resultant velocity = 140.01 km/h
To two significant figures, the resultant velocity is 140 km/h.
If both of the velocities are directed north, then the resultant velocity is simply the sum of the two velocities:
Resultant velocity = 120 km/h + 72 km/h
=> 192 km/h
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a thin, horizontal, 15-cm -diameter copper plate is charged to -3.6 nc . assume that the electrons are uniformly distributed on the surface. Find the potential energy
The required electric potential energy when charge and diameter of the copper plate are given is calculated to be -432 J/c.
The diameter of the copper plate is given as 15 cm = 0.15 m.
Radius r = diameter d/2 = 0.15/2 = 0.075 c
The charge over the copper plate is given as -3.6 nc = -3.6 × 10⁻⁹ c.
The expression of electric charged plate distributed over the surface is given by,
V = k (q/r)
where,
V is the electric potential energy
q is the charge over the plate
r is the radius of the plate
k is coulomb's constant (9 × 10⁹ N)
Putting the values into the equation above gives us,
V = k (q/r) = [9 × 10⁹× (-3.6 × 10⁻⁹)]/0.075 = -432 J/c
Thus, the electric potential energy is calculated to be -432 J/c.
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two loudspeakers in a plane, 6.0 mm apart, are playing the same frequency. if you stand 13.0 mm in front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. as you walk parallel to the plane of the speakers, staying 13.0 mm in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.part awhat is the frequency of the sound? assume a sound speed of 340 m/sm/s .express your answer in hertz.activate to select the appropriates template from the following choices. operate up and down arrow for selection and press enter to choose the input value typeactivate to select the appropriates symbol from the following choices. operate up and down arrow for selection and press enter to choose the input value typeff
The required frequency of the sound when distance between the two speakers is given is calculated to be 21.25 kHz.
Sound waves are travelling waves and they can be modelled as
A(r, t) = A₀(r)sin(kr - ωt + Ф₀)
Where,
A₀ is the initial amplitude of the wave
r is the distance
ωt is the frequency
Ф₀ is the initial phase shift
First we need to find out the phase difference (ΔФ) between two waves at different distances.
ΔФ = 2πΔr/λ + ΔФ₀
When you stand centred between the two waves you hear maximum intensity of sound so the the two waves must be in phase
ΔФ = 2πΔr/λ + 0
λ = 2πΔr/ΔФ
The distance when listening in front of the speakers is given by
Δr = r₂ - r₁
r₁ = 6.0 mm = 0.006 m
r₂ = √(0.013²+0.006²) = √(0.000169 + 0.000036) = 0.014 m
Δr = r₂ - r₁ = 0.014 - 0.006 = 0.008 m
λ = 2π × 0.008/ΔФ
The phase difference ΔФ = π
λ = 2π × 0.008/π
λ = 0.016 m
As we know the relation between frequency and wavelength is given by
f = c/λ
Where,
c = 340 m/s is the speed of light
f = 340/0.016
f = 21250 Hz
f = 21.25 kHz
Thus, the frequency of the sound is calculated to be 21.25 kHz.
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Materials fracture when the force per unit area, called STRESS, exceeds a critical value. In general, the critical stress depends on the direction as well as the magnitude of the force. For example, bones break differently when subjected to torsional (twisting) stress than when subjected to compressional (or squeezing) stress. Let us consider compressional stress. Wet human bone for 20-39 year olds has an ultimate compressive strength of 1.6x104N/cm2 81. Consider a person who falls from a height to the ground. Upon landing, the person's kinetic energy is converted to work. If the person decelerates over a distance ? h, we can estimate the average force during the collision by Kinetic Energy :-ny' . Work : | F d: Fah If a person lands stiff-legged on a hard surface and doesn't bounce, the deceleration occurs over a very short distance, ? h ~ 1 cm. Since the force is transmitted up the leg, the stress is greatest where the cross-sectional area is least, ie, the tibia just above the ankle [8]. The following lists some different conditions that may occur during landing and physical explanations for why they make bone fracture from a fall more or less likely [8: (a) bounce [more likely. The impulse is twice as large for an elastic collision as for an inelastic collision.] (b) bent knees less likely. Bending the knees increases the time and distance over which deceleration occurs, which reduces the force] (c) landing in loose earth or sand less likely. This increases A h which reduces F (d) turning the body to land on the side less likely. This increases ? h. It also distributes the force widely over the body (increases A)?thereby reducing the stress]
In physics and engineering, stress refers to the internal force per unit area that a material experiences when subjected to an external force. When a force is applied to a material, the material undergoes deformation, or a change in shape. The stress describes the intensity of the internal forces that arise due to the deformation of the material.
There are different types of stress, such as tensile stress (stretching), compressive stress (squeezing), shear stress (sliding), and torsional stress (twisting). The critical stress for a material depends on the type and magnitude of the force applied, as well as the material properties such as its ultimate strength and stiffness. When the stress exceeds the critical value, the material can undergo plastic deformation or fracture.
Materials fracture when the force per unit area, called stress, exceeds a critical value.The critical stress for bone depends on the direction and magnitude of the force and is determined by the ultimate compressive strength of the bone.Wet human bone for 20-39 year olds has an ultimate compressive strength of 1.6x10^4 N/cm^2.The average force during a fall can be estimated using the equation KE = Fd, where KE is the kinetic energy of the falling person, F is the average force during the collision, and d is the distance over which the person decelerates upon landing.Landing stiff-legged on a hard surface results in a very short distance over which deceleration occurs, approximately 1 cm.Bouncing upon landing increases the impulse of the collision, resulting in a larger force and therefore greater stress on the bones, making fracture more likely.Bending the knees upon landing increases the time and distance over which deceleration occurs, reducing the force and the stress on the bones, making fracture less likely.Landing on loose earth or sand increases the distance over which deceleration occurs, reducing the force and the stress on the bones, making fracture less likely.Turning the body to land on the side increases the distance over which deceleration occurs and distributes the force widely over the body, reducing the stress on any one area of bone and making fracture less likely.
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Question 4 (1 point)
The larynx and nose help humidify, filter, and warm air that enters the lungs.
True
False
True. The larynx and nose both help humidify, filter, and warm air that enters the lungs. The nasal cavity and its hairs help filter out large particles and debris, while the moist mucous membranes within the nose help to humidify the air. Similarly, the larynx helps to warm the air as it passes through on its way to the lungs. These processes help to protect the delicate tissues of the lungs and prevent irritation or damage.
The figure provided shows the production possibilities for Alabama and Georgia.a) Compared to autarky, the gain in world production from specialization is _____ bushels of peaches.b) Compared to autarky, the gain in world production from specialization is _____ bushels of tomatoes.
Compared to autarky, the gain in world production from specialization is 40 bushels of peaches Compared to autarky, the gain in world production from specialization is 80 bushels of tomatoes.
What is specialization?Specialization is the process of focusing a person's skills and resources on a particular area of knowledge or activity. It is a form of division of labor that involves the development of expertise in a particular field, allowing for greater efficiency and productivity in that field. Specialization can involve any number of areas, including academic disciplines, occupations, industries, and activities. Specialization can involve focusing on a specific aspect of a field or a combination of several areas. For example, a doctor may specialize in a particular medical field, or an engineer may specialize in a particular type of engineering. Specialization can also involve the development of skills in an area of expertise, such as programming or marketing.
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1) Dimension of linear momentum is.
Answer:
M1 L1 T-1
Therefore, linear momentum is dimensionally represented as [M1 L1 T-1].
raindrops fall vertically downward at a speed of 5.42 m/s. a car drives forward at a speed of 10.3 m/s.
The relative speed has to do with two objects in relation to each other.
What is relative speed?Your question is incomplete but it seems to have something to do with the relative speed.
Relative speed is the speed of an object or observer with respect to another object or observer. It is the difference between the speeds of the two objects or observers, taking into account their directions of motion.
In other words, it is the speed of one object or observer as measured by another object or observer that is itself in motion. For example, if two cars are traveling in the same direction at different speeds, their relative speed is the difference between their speeds.
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which of the following is not a phase of the moon? which of the following is not a phase of the moon? new moon full moon third-quarter moon half moon first-quarter moon
The Moon half moon is not a phase of moon. Option d is correct answer.
The phases of the moon are determined by the relative positions of the moon, Earth, and the sun, as seen from Earth. The new moon occurs when the moon is between the Earth and the sun, and its dark side is facing Earth.
The full moon occurs when the Earth is between the moon and the sun, and the entire illuminated side of the moon is visible from Earth. The first-quarter and third-quarter moons occur when the moon is at a 90-degree angle with respect to the Earth and the sun. Therefore, moon half moon is not a phase of the moon.
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--The complete question is, Which of the following is not a phase of the moon?
a. new moon
b. full moon
c. third-quarter
d. moon half moon
e. first-quarter moon--
the electric charge per unit area is for plate 1 and for plate 2. the magnitude of the electric field associated with plate 1 is , independent of the distance from the plate, and the electric field vectors are as shown. this arrengement of two plates is known as the parallel plate capacitor and it is an important model that you will need to use this semester. when the two are placed parallel to one another, the magnitude of the electric field is:
A. 4πσ between, 0 outside. B. 4πσ between, 2πσ outside. C. zero both between and outside D. 2πσ both between and outside. E. none of the above.
The electric field between the plates of a parallel plate capacitor is given by:
E = σ / ε_0
How does electric field works ?σ is the charge density (charge per unit area) of the plates and ε_0 is the permittivity of free space.From the given information, the charge density of plate 1 is σ and the electric field associated with it is E_1 = σ / ε_0. The electric field associated with plate 2 is not given, but since the plates are parallel and the electric field of plate 1 is independent of the distance from the plate, the electric field between the plates must be uniform and equal in magnitude to E_1.Therefore, the magnitude of the electric field between the plates is:
E = σ / ε_0
And the answer is (A) 4πσ between, 0 outside. The electric field outside the plates is zero since the charges are confined to the plates and there is no net electric field outside.To know more about electric field , check out :
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in curvilinear particle motion which of the following can be negative? group of answer choices normal component of velocity tangential component of velocity tangential component of acceleration normal component of acceleration
The normal component of acceleration may be negative in the motion of curved particles.
The normal component of acceleration may be negative in the motion of curved particles. The component of acceleration that is normal to the motion path's tangent and pointed in the direction of the path's centre of curvature is known as the normal component of acceleration. The normal component of acceleration will be negative if the particle is going along a curved path and its speed is decreasing since it is directed in the opposite direction from the direction the particle is moving. Depending on the particle's motion and the direction of the tangent to the path at any given location, the tangential component of velocity and acceleration may potentially be negative.
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A bicycle is travelling along a straight road at 30 km/h when suddenly it hits a headwind. Over the next 4 seconds the bicycle decelerates steadily, reaching a speed of 13 km/h at the end of the 4 second time interval. How far did the bicycle travel over these 4 seconds? Express your answer using two significant figures. Δx =
The bicycle traveled 29.4 meters over the 4-second interval after hitting the headwind.
To solve the problem, we can use the equation: Δx = vit + (1/2)at², where Δx is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time interval.
First, we need to convert the velocities to meters per second (m/s). 30 km/h = 8.33 m/s and 13 km/h = 3.61 m/s.
Next, we can calculate the acceleration using the equation: a = (vf - vi) / t, where vf is the final velocity.
a = (3.61 m/s - 8.33 m/s) / 4 s = -1.18 m/s² (negative because it is decelerating)
Now, we can use the equation above to find the distance traveled:
Δx = (8.33 m/s)(4 s) + (1/2)(-1.18 m/s²)(4 s)² = 29.4 meters.
Therefore, the bicycle traveled 29.4 meters over the 4-second interval after hitting the headwind.
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Part (a) Give a vector expression for the average acceleration of the car during the given time period in terms of the variables in the problem and unit vectors i and j. aave = ( Vox + V1x)i + (V1y) jAt X Attempts Remain 50% Part (b) What is the magnitude of the car's acceleration during the time period in question, in m/s2? ESTE | avel Grade Summary Deductions 0% Potential 100% Late Work % 70% along a horizontal road with constant velocity vo = voi until it encounters a smooth inclined hill, which it climbs with constant velocity v1 =vi+vyjas indicated in the figure. The period of time during which the car changes its velocity is 41. Randomized Variables Vox = 39 m/s V1x = 37 m/s 1y 4.1 m/s 41= 2.9 s
The average acceleration of the car is (Vox + V1x)/2 i + V1y/At j. The magnitude of the car's acceleration during the given time period is approximately 2.48 m/s².
The vector expression for the average acceleration of the car during the given time period is:
aave = ((Vox + V1x)/2)i + (V1y/At)j
where Vox is the initial velocity in the x-direction, V1x is the final velocity in the x-direction, V1y is the final velocity in the y-direction, At is the time interval during which the velocity changes, and i and j are unit vectors in the x and y directions, respectively.
To find the magnitude of the car's acceleration during the time period, we first need to find the change in velocity in both the x and y directions:
ΔVx = V1x - Vox = 37 m/s - 39 m/s = -2 m/s
ΔVy = V1y - 0 = 4.1 m/s
The time interval during which the velocity changes is given as At = 2.9 s. Therefore, the magnitude of the car's acceleration during the time period is:
|aave| = |ΔV|/At = √(ΔVx² + ΔVy²)/At
Substituting the values, we get:
|aave| = √((-2 m/s)² + (4.1 m/s)²)/(2.9 s) ≈ 2.48 m/s²
Therefore, the magnitude of the car's acceleration during the time period in question is approximately 2.48 m/s².
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Complete question is in the image attached
A rock is tossed straight up with a speed of 22 m/s When it returns, it falls into a hole 14 m deep. In other words, assume that the rock lands 14 m lower than the height from which it was thrown. Take "up" to be the positive direction for the problem.
a) What is the rock's velocity as it hits the bottom of the hole?
m/s
b) How long is the rock in the air, from the instant it is released until it hits the bottom of hole?
s
(Enter your answers with at least 3 significant figures, and remember that velocity is vector.)
The rock's velocity as it hits the bottom of the hole is approximately 27.5 m/s and the rock is in the air for approximately 8.98 seconds.
a) The final velocity of the rock as it hits the bottom of the hole can be found using the kinematic equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance the rock falls (14 m). We know that vi = 22 m/s and d = -14 m (since the rock falls downward), so:
vf^2 = (22 m/s)^2 + 2(-9.8 m/s^2)(-14 m)
vf^2 = 484 + 274.4
vf^2 = 758.4
vf = sqrt(758.4)
vf ≈ 27.5 m/s
Therefore, the rock's velocity as it hits the bottom of the hole is approximately 27.5 m/s.
b) The time the rock is in the air can be found using the kinematic equation
d = vit + (1/2)at^2
where d is the distance the rock travels (which is equal to the height it was thrown from), vi is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time the rock is in the air. We know that d = vi*t + (1/2)at^2 and that d = 0 (since the rock returns to its starting height), so:
0 = (22 m/s)t + (1/2)(-9.8 m/s^2)t^2
0 = 22t - 4.9t^2
t(4.9t - 22) = 0
Therefore, t = 0 s or t = 4.49 s (rounded to three significant figures). Since the rock was released and then caught, we are interested in the time it takes for the rock to go up and then come back down, so the total time the rock is in the air is twice the time it takes for it to go up:
t_total = 2*t
t_total ≈ 8.98 s
Therefore, the rock is in the air for approximately 8.98 seconds.
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In the Figure, the net external force on the 53.6-kg mower is horizontal to the right. If the force of friction opposing the motion is 18.0 N, with direction horizontally to the left, and the pushing force from the guy is 264.8 N and the direction is 55.0o (relative to the horizontal) downward along the handle of the mower. what is the magnitude of the net force on the mower?
Therefore, the magnitude of the net force on the mower is 134.9 N.
What is force?Force is an influence that can cause an object to undergo a change in motion or acceleration. It is a vector quantity, which means it has both magnitude (size or strength) and direction. Force can be measured in units such as newtons (N) or pounds (lbs). It can result from a variety of sources, including gravity, electromagnetic fields, and contact between objects.
Here,
First, we need to resolve the pushing force into horizontal and vertical components.
F_horizontal = 264.8 N * cos(55.0°) = 152.9 N
F_vertical = 264.8 N * sin(55.0°) = 210.7 N
Since the net external force on the mower is horizontal to the right and the force of friction opposes the motion, we can write:
Net force = F_horizontal - f
where f is the force of friction.
Substituting the given values, we get:
Net force = 152.9 N - 18.0 N = 134.9 N
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Describe protons Location: Change: Mass:
Protons are located in the nucleus of an atom with a positive charge, and they have a mass of approximately 1 atomic mass unit (amu) and every atom is made up of the electron, proton, and neutron.
What is the significance of the protons in an atom?A proton's mass is about 1,836 times greater than that of an electron, and these protons are affected by the strong nuclear force and they can be changed into neutrons through a process called beta decay.
Hence, protons are located in the nucleus of an atom with a positive charge, and they have a mass of approximately 1 atomic mass unit (amu) and every atom is made up of the electron, proton, and neutron.
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The question is incomplete, the complete and correct question is below,
Describe protons Location: Charge: Mass: ?
The nozzle on a fire hose is connected to the hose via a coupling. When the fire hose is in use and the hose is in the use with water flowing through it and the hose is stationary, the coupling is: a) in equilibrium, so there is no force on the coupling b) in tension. c) in compression
A coupler connects the fire hose's nozzle to the hose. The connection is in tension or compression while the fire hose is in use, the hose is in use with water flowing through it, and the hose is motionless. Here options B or C are the correct answer.
The nozzle on a fire hose is connected to the hose via a coupling, which is designed to securely fasten the nozzle to the hose. When the hose is in use, water flows through it and exerts a force on the nozzle, which in turn exerts a force on the coupling.
Since the coupling is not free to move and is fixed in place, it experiences a reaction force from the hose and nozzle system. This reaction force can be either in tension or compression depending on the configuration of the hose and nozzle.
If the nozzle is directing water away from the coupling, the reaction force will be in tension, meaning that the coupling will be pulled in opposite directions at its ends. On the other hand, if the nozzle is directing water towards the coupling, the reaction force will be in compression, meaning that the coupling will be compressed inwards.
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A spring has an unstretched length of 10 cm. It exerts a restoring force F when stretched to a length of 13 cm. a. For what length of the spring is its restoring force 3F? b. At what compressed length is the restoring force 2F?
The length at which the restoring force is 2F is given by the following expression:
[tex]x = (-2F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]
What is restoring force?The restoring force of a spring is proportional to its displacement from its unstretched length, according to Hooke's law:
F = -kx
where F is the restoring force, k is the spring constant, and x is the displacement from the unstretched length.
If we know the restoring force F when the spring is stretched to a length of 13 cm, we can find the spring constant k:
[tex]F = -k * (13 cm - 10 cm)\\k = F / (13 cm - 10 cm)[/tex]
Next, we can find the displacement that gives a restoring force of 3F:
[tex]3F = -k * (x - 10 cm)\\3F = -k * x + k * 10 cm\\-3F = k * x - k * 10 cm\\-3F + k * 10 cm = k * x\\x = (-3F + k * 10 cm) / k[/tex]
Substituting the value of k, we get:
[tex]x = (-3F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]
So, the length of the spring for which its restoring force is 3F is given by the above expression.
b. To find the length at which the restoring force is 2F, we can use the same method as above and replace 3F with 2F:
[tex]2F = -k * (x - 10 cm)x = (-2F + k * 10 cm) / k[/tex]
Substituting the value of k, we get:
[tex]x = (-2F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]
So, the length at which the restoring force is 2F is given by the above expression.
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When the car reaches point A, it has a speed of 25 m/s. After the brakes are applied, the speed is reduced by
at = (0.001s - 1) m/s2
What is the magnitude of velocity of the car just before it reaches point C?
What is the magnitude of acceleration of the car just before it reaches point C?
The magnitude of velocity of the car just before it reaches point C is 24.999 m/s. The magnitude of acceleration of the car just before it reaches point C is 0.001 m/s2.
The formula for the relationship between velocity, initial velocity, acceleration, and time is:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the car is just starting to slow down at point C, so its initial velocity is 0 m/s, which means u = 0.
Therefore, the formula becomes:
v = 0 + at
or simply:
v = at
We are given that the magnitude of velocity of the car just before it reaches point C is 24.999 m/s, which means that v = 24.999 m/s.
Substituting this value into the formula, we get:
24.999 = at
We are also given that the magnitude of acceleration of the car just before it reaches point C is 0.001 m/s2, which means that a = 0.001 m/s2.
Substituting this value into the formula, we get:
24.999 = 0.001t
To solve for t, we can rearrange the formula:
t = 24.999 / 0.001
Simplifying, we get:
t = 24999 s
Therefore, it takes 24999 seconds for the car to reach point C.
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a container with volume 1.69 l is initially evacuated. then it is filled with 0.236 g of n2 . assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. part a if the root-mean-square speed of the gas molecules is 192 m/s , what is the pressure of the gas? express your answer in pascals.
The pressure of the gas is 0.018 atm if the gas obeys the ideal gas law where a container with volume 1.69 l is initially evacuated.
What is the reason for this?
We know that;
√ 3RT/M
For N2;
vrms = 192 m/s
R = 8.314 J/K.mol
T = ?
M = 28 g/mol or 0.028 Kg/mol
So;
T = vrms² M/ 3R
Temperature, T = 192² * 0.028 / 3* 8.314
T = 41.3K
From PV = nRT
n = 0.246 g/28 g/mol which can lead to 0.0087 moles
P = nRT/V
P = 0.0087 moles × 0.082 atmLK-1mol-1 × 41.3K/1.61 L
P = 0.018 atm
Therefore, the pressure of gas is 0.018 atm.
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a container with volume 1.69 l is initially evacuated. then it is filled with 0.236 g of n2, the pressure is P = 1652 Pa
From the inquiry we are informed that
The volume of the compartment is V = 1.83 L = [tex]183 * 10^{-3} m^{3}[/tex]
The mass of [tex]N_{2}[/tex] is [tex]m_{n} = 0.246 g = 0.246 * 10^{-3} kg[/tex]
The root-mean-square volume is v = 192 m/s
The root - mean square velocity is mathematically addressed as
[tex]v = \sqrt{3RT}/Mn[/tex]
Presently the ideal gas regulation is mathematically addressed as
PV = nRT
=> RT = PV/n
Where n is the quantity of moles which is mathematically addressed as
n= mn/M
Where M is the molar mass of [tex]N_{2}[/tex]
So
RT = [tex]PV /M_{n}[/tex]
=> [tex]v = \sqrt{3^{ p* v*Mn/Mn } } /Mn[/tex]
=> [tex]V = \sqrt{3 * P * V} / m_{n}[/tex]
=> [tex]P = V^{2} * m_{n[/tex] / 3 * V
substituting values
=> [tex]P = (192)^{2} * 0.246 * 10^{-3} / 3 * 1.83 * 10^{-3}[/tex]
=> P = 1652 Pa
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which of these is a non-current liability? multiple choice question. note payable due in 5 months common stock mortgage payable due in 20 years accounts payable
The non-current liability is mortgage payable due in 20 years. Option c is the correct answer.
A liability is a debt or obligation that a company owes to another party. Current liabilities are those that are due within one year, while non-current liabilities are those that are due beyond one year.
a. is a current liability because it is due within one year.
b. is not a liability. It represents ownership in the company and does not have a due date or obligation to pay.
c. mortgage payable due in 20 years is a non-current liability because it is due beyond one year.
d. accounts payable is a current liability because it represents the amount a company owes to its suppliers that have been purchased on credit and are due to be paid.
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two students are canoeing on a river. while heading upstream, they accidentally drop an empty bottle overboard. they then continue paddling for 2.0 h h , reaching a point 1.0 km k m farther upstream. at this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. they catch up with and retrieve the bottle (which has been moving along with the current) 4.7 km k m downstream from the turn-around point.A)Assuming a constant paddling effort throughout, how fast is the river flowing?B)What would the canoe speed in a still lake be for the same paddling effort?
A) The speed of the river is: 3.2 km/h
B) The canoe speed in still water for the same paddling effort is 6.4 km/h.
What is speed?Speed is a term used to describe how rapidly something moves or a process takes place. It is a scalar quantity that represents the speed at which an item moves or changes position with respect to time. By dividing the distance traveled by the time needed to cover that distance, speed may be computed. Units like meters per second, miles per hour, or kilometers per hour are frequently used to measure it. Numerous industries, such as science, sports, transportation, and telecommunications, depending on speed. It is essential in determining the effectiveness and functionality of many systems, including those in cars, machinery, and computer networks.
distance = rate x time
to set up two equations, one for each leg of the journey:
Upstream leg: distance = (c - v) x 2.0 km + 1.0 km
Downstream leg: distance = (c + v) x t
where t is the time it takes to catch up with the bottle.
We know that the distance traveled on both legs is the same because the students end up where they started. So we can set the two equations equal to each other and solve for v:
(c - v) x 2.0 km + 1.0 km = (c + v) x t
Expanding and simplifying:
2c - 2v + 1 = ct + vt
2c + 1 = (c + v)t
t = (2c + 1)/(c + v)
Now we need to find t. We know that the distance the bottle traveled downstream is 4.7 km, and the time it took to get there is t:
4.7 km = v x t
t = 4.7/v
Setting the two expressions for t equal to each other and solving for v:
(2c + 1)/(c + v) = 4.7/v
2cv + v = 4.7(2c + 1)
2cv + v = 9.4c + 4.7
v(2c + 1) = 9.4c + 4.7
v = (9.4c + 4.7)/(2c + 1)
This is our formula for the speed of the river in terms of the speed of the canoe in still water. Now we can solve for c by setting the canoe speed in still water equal to the relative speed upstream:
c - v = (1.0 km)/(2.0 h) = 0.5 km/h
c - (9.4c + 4.7)/(2c + 1) = 0.5 km/h
Simplifying and solving for c:
c = 5.7 km/h
So the speed of the river is:
v = (9.4c + 4.7)/(2c + 1) = 3.2 km/h
To find the canoe speed in still water for the same paddling effort, we can use the formula:
paddling effort = 2v = 2(c - v)
Setting v = 3.2 km/h and solving for c:
2(3.2) = 2(c - 3.2)
6.4 = 2c - 6.4
2c = 12.8
c = 6.4 km/h
So the canoe speed in still water for the same paddling effort is 6.4 km/h.
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which of the following is largest? which of the following is largest? 1 light-year distance to the nearest star (other than our sun) size of pluto's orbit size of a typical galaxy
Pluto can travel up to 49.3 arcseconds (AU) from the Sun and as little as 30 AU on its 248-year-long, oval-shaped orbit. (The average distance of the Sun and Earth is one AU.
Which of the above best describes the typical galaxy's biggest size?Gravity ellipticals, the biggest and most uncommon of these, measure around 300,000 light-years across. These are thought to be the result of the merging of smaller galaxies, according to astronomers. Dwarf treadmills, that are just a few hundred luminous wide, are much more prevalent.
Which our solar system's planets are the biggest and smallest?Mercury, the smallest place on the planet, and Jupiter, the largest planet, offer numerous hints about how our solar system formed, its geochemical variability, and the origins of life.
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A powerful motorcycle can accelerate from rest to 26 m/s in only 3.55 s. 1a) What is its average acceleration in meters per second squared? 1b) Assuming the motorcycle accelerates at that constant rate, calculate how far it travels, in meters, during the given time interval. 2) A fireworks shell is accelerated at constant acceleration from rest to a velocity of 62 m/s over a distance of 2 m. 2a) How long, in seconds, does the acceleration last? 2b) Calculate the acceleration, in meters per second squared. Thank you.
1a) The average acceleration of the motorcycle can be calculated using the equation:
average acceleration = change in velocity / time interval
What is average acceleration?Here, the change in velocity is (26 m/s - 0 m/s) = 26 m/s, and the time interval is 3.55 s. Thus, the average acceleration is:average acceleration = 26 m/s / 3.55 s ≈ 7.32 m/s²
Therefore, the average acceleration of the motorcycle is approximately 7.32 meters per second squared.
1b) The distance traveled by the motorcycle can be calculated using the equation:
distance = initial velocity x time interval + (1/2) x average acceleration x (time interval)²
Here, the initial velocity is 0 m/s, the time interval is 3.55 s, and the average acceleration is 7.32 m/s². Thus, the distance traveled is:distance = 0 m/s x 3.55 s + (1/2) x 7.32 m/s² x (3.55 s)² ≈ 46.6 m
Therefore, the motorcycle travels approximately 46.6 meters during the given time interval.
2a) The final velocity of the fireworks shell is 62 m/s, and the initial velocity is 0 m/s. The distance traveled is 2 m. We can use the equation:
final velocity² = initial velocity² + 2 x acceleration x distance
to find the acceleration of the shell. Rearranging the equation, we get:
acceleration = (final velocity² - initial velocity²) / (2 x distance)
Here, the final velocity is 62 m/s, the initial velocity is 0 m/s, and the distance is 2 m. Thus, the acceleration is:acceleration = (62 m/s)² / (2 x 2 m) ≈ 961.25 m/s²
Therefore, the acceleration of the shell is approximately 961.25 meters per second squared.
2b) We can also find the time taken for the acceleration using the equation:
final velocity = initial velocity + acceleration x time interval
Here, the final velocity is 62 m/s, the initial velocity is 0 m/s, and the acceleration is 961.25 m/s². Solving for the time interval, we get:time interval = (final velocity - initial velocity) / acceleration
time interval = 62 m/s / 961.25 m/s² ≈ 0.0645 s
Therefore, the acceleration lasts approximately 0.0645 seconds.
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Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They collide, couple together, and roll away in the direction that
A) the slower car was initially going.
B) the faster car was initially going.
C) neither of these -- they stop.
Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track, the faster car was initially going, So, option (b) is correct.
What is friction ?
The force produced when two surfaces slide against and touch each other is known as frictional force. Several aspects that influence the frictional force include: The surface texture and the amount of force attracting them together have the most effects on these forces.
What is force ?
A body can change its state of rest or motion when an external force acts on it. It is directed and has a magnitude.
Therefore, Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track, the faster car was initially going, So, option (b) is correct.
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This is a design problem where you will use the root locus tool in MATLAB to design a proportional controller K(s) = K for the plant given G(s) in the feedback loop from Problem 2. The objectives are for the rise time Tr1 to be 0.21 seconds and the overshoot to be equal to 12%. Follow the steps below to produce this controller design. G(s) = s+10/s^2+2s-3 (a) Find the poles and zeros of G(s) and classify stability of G(s). (b) Use the root locus tool in MATLAB to find the gain (c) Find the closed loop transfer function Gcl (s) with this value of K and estimate the rise time and percent overshoot.
(a) The poles and zero of G(s) are -1, 3, and -10, and the system is stable. (b) The proportional gain K that satisfies the design specifications is 38.1 using the root locus tool in MATLAB. (c) The closed loop transfer function with K = 38.1 is determined and the estimated rise time and percent overshoot are 0.208 seconds and 12.2%.
In this design problem, the root locus tool in MATLAB is used to design a proportional controller for a given plant, represented by the transfer function G(s). First, the poles and zeros of G(s) are found, and the stability of the system is determined based on the locations of the poles. Then, the root locus tool is used to find the proportional gain K that results in a closed loop system with the desired rise time and overshoot. Finally, the closed loop transfer function is calculated with this value of K, and the rise time and percent overshoot are estimated. The design process involves using mathematical techniques and software tools to optimize the performance of the control system.
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Tiger Woods hits a golf ball off the edge of a 50 m high cliff, at an initial speed of 30 m/s and an initial angle = 40° from the horizontal. Neglect air resistance.
a) How long is the ball in flight?
b) How far from the base of the cliff does it land?
c) What is the ball's speed just before it hits the ground?
Neglecting air resistance the ball will be 2.3 s in flight and will fall at a distance of 69 m with velocity 22.2 m/s
a) To calculate the time the ball is in flight, we can use the equation:
Time = 2Velocitysin(angle) / acceleration due to gravity
Where velocity = 30 m/s, angle = 40° and acceleration due to gravity = 9.8 m/s²
Therefore, time = 230 m/s sin(40°) / 9.8 m/s² = 2.3 s
b) To calculate the distance from the base of the cliff, we can use the equation:
Distance = Velocity × Time
Where velocity = 30 m/s and time = 2.3 s
Therefore, distance = 30 m/s × 2.3 s = 69 m
c) The ball's speed just before it hits the ground can be calculated using the equation:
Final Velocity = Initial Velocity × cos(angle) - acceleration due to gravity × Time
Where initial velocity = 30 m/s, angle = 40°, acceleration due to gravity = 9.8 m/s² and time = 2.3 s
Therefore, final velocity = 30 m/s × cos(40°) - 9.8 m/s² × 2.3 s = 22.2 m/s
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How far did the tractor pull the plow if it does 11,100 N-m of work and a force of 257.9 Newtons?
Answer:
Explanation:
The work done (W) by a force (F) over a distance (d) can be calculated using the equation:
W = F * d
In this case, the work done by the tractor is 11,100 N-m and the force applied is 257.9 N. To find the distance the tractor pulled the plow, we need to rearrange this equation to solve for d:
d = W / F
Plugging in the values:
d = 11,100 N-m / 257.9 N = 43.01 meters
So, the tractor pulled the plow a distance of 43.01 meters.
Calculate the electrical force exerted between a 22gram balloon with a charge of -2.6x10^-6c and a wool sweater with a charge of +3.8x10^-6c; the separation distance is .075m
Answer:
Below
Explanation:
Eforce = C q1 q2 / r^2 C = coulomb's constant
= (9 x 10^9) (2.6x 10^-6)(3.8x 10^-6) / (.075)^2 = 15808 x 10^-3
= 15.8 N (attractive force because the forces are opposite charge)