Answer:
Q = 12.5 kJ
Explanation:
The expression to use to calculate Heat is:
Q = H° * n
Where:
Q: heat (J or kJ)
H°: enthalpy of reaction (kJ/mol)
n: moles
Now, as it was stated in the comments, the question is incomplete, and here is the missing part:
Given:
2A + B A2B (1)
H° = – 25.0 kJ/mol
2A2B 2AB + A2 (2)
H° = 35.0 kJ/mol
With these two reactions, we can calculate the heat.
Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:
2A + B -------> A2B H°1 = -25 kJ/mol
2A2B --------> 2AB + A2 H°2 = 35 kJ/mol
Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:
2A + B + 2A2B -------> 2AB + A2
And the enthalpy, it's just summed:
H°3 = -25 + 35 = 10 kJ/mol
Now with this value we can calculate heat:
Q = 10 * 2.5 = 25 kJ
However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:
Q = 25 / 2 = 12.5 kJ
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out.
To calculate the heat required for the reaction, we need to know the enthalpy change (ΔH°) for the reaction 2a + b → a2b. However, the enthalpy change for this reaction is not provided in the question. Assuming that the enthalpy change for this reaction is known, we can proceed with the calculation.
Let's denote the enthalpy change for the reaction 2a + b → a2b as ΔH°. The reaction of interest is:
[tex]\[ 2a + b + a_2b \rightarrow 2ab + a_2 \][/tex]
This reaction can be broken down into two parts:
1. The reaction of 2 moles of a with 1 mole of b to form 1 mole of a2b:
with an enthalpy change of H°.
[tex]\[ 2a + b \rightarrow a_2b \][/tex]
2. The reaction of 1 mole of a2b to form 2 moles of ab:
[tex]\[ a_2b \rightarrow 2ab \][/tex]
Since the reaction of a2b to form 2ab does not involve any additional reactants, it can be considered as the reverse of the first part, but with twice the amount of ab produced. Therefore, the enthalpy change for this part would be -H° (since the reverse reaction has the opposite sign of the forward reaction), and for 1 mole of a2b reacting, it would be -2ΔH° (because 2 moles of ab are formed).
Now, we are given 2.50 moles of a, and we assume there is excess b and a2b. The reaction will proceed until all of a is consumed. Since the reaction stoichiometry is 2 moles of a to 1 mole of a2b, 1.25 moles of a2b will be required to react with the 2.50 moles of a (since 2.50 moles of a require 1.25 moles of a2b).
The overall enthalpy change for the reaction will be the sum of the enthalpy changes for the two parts:
1. For the reaction of 2.50 moles of a with b to form a2b:
[tex]\[ \Delta H_1 = 2.50 \text{ moles} \times \Delta H^\circ \][/tex]
2. For the reaction of 1.25 moles of a2b to form ab:
[tex]\[ \Delta H_2 = 1.25 \text{ moles} \times (-2 \Delta H^\circ) \][/tex]
The total enthalpy change (ΔH_total) is the sum of ΔH_1 and ΔH_2:
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex][tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2 \times 1.25 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2.50 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 0 \][/tex]
Surprisingly, the total enthalpy change for the reaction is zero. This is because the amount of heat released in the formation of a2b is exactly equal to the amount of heat absorbed in the formation of ab from a2b. Therefore, no additional heat is required for the overall reaction to proceed.
The final answer is: [tex]\[ \boxed{0} \][/tex]
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out. This conclusion is based on the assumption that the enthalpy change for the reaction 2a + b → a2b is known and that the reaction proceeds as written. If the enthalpy change for the formation of a2b is not known, then it would be necessary to look up the standard enthalpy of formation for a2b to perform the calculation.
A solar eclipse will occur Group of answer choices
1. at every full Moon. during every new Moon.
2. at full Moon only when the Moon is on the ecliptic.
3. at new Moon only when the Moon is on the ecliptic.
Answer:
3. at new Moon only when the Moon is on the ecliptic.
Explanation:
Solar eclipse is the condition when the moon comes in between the sun and the earth. In this condition the moon casts its shadow on the earth.Whether the eclipse is a total solar eclipse, a partial solar eclipse or an annular solar eclipse depends on various factors, but the position of the moon must be on the same orbital plane as that of the earth's orbit around the sun.The sun is about 400 times larger than the moon in size and the sun is almost 400 times farther from the earth than the moon is, this makes it possible for the moon to cover the sun completely leading to a complete solar eclipse.As we know that the orbit of the earth around the sun and the orbit of the moon around the earth is elliptical which leads to a variation in the distance from their rotating centers, so not of every eclipse the moon covers the sun completely developing an annular eclipse.When the moon is close enough to the earth on the ecliptic but not completely aligned in between the sun and the earth leads to a partial solar eclipse.A solar eclipse occurs during a new Moon, exclusively when the Moon is on the ecliptic path. This is when the Moon moves between the Earth and the Sun and blocks the sunlight. A solar eclipse can't occur during a full Moon.
Explanation:A solar eclipse happens during a new Moon when the Moon is on the ecliptic path. This event occurs when the Moon moves between the Earth and the Sun and blocks the Sun's light. The Moon must be on the ecliptic, which is the apparent path of the Sun's motion on the celestial sphere, for the alignment to be precise. If the Moon is not on this path, then it will not block the Sun, and a solar eclipse will not occur. It's also important to note that during a full Moon, the Moon is opposite the Sun in our sky, so a solar eclipse cannot occur.
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Please help, it’s much needed
Answer:
Displacement = 72m
Acceleration = 4 [tex]m/s^{2}[/tex]
Explanation:
The jet ski increases its speed from zero to 24 m/s in 6 seconds.
We are supposed to find the displacement and its acceleration rate.
Displacement is the distance it covers in this time interval.
Acceleration is the rate of increase of speed.
Acceleration = [tex]\frac{change in velocity}{time}[/tex]
Acceleration = [tex]\frac{24 - 0}{6}[/tex]
Acceleration = 4 [tex]m/s^{2}[/tex]
According to equation of motion ,
[tex]v^{2}-u^{2} = 2\times a\times s[/tex]
[tex]24^{2} - 0 = 2\times 4\times s[/tex]
s = [tex]\frac{576}{8}[/tex]
Displacement , s = 72 m
Which layer of the atmosphere absorbs the most ultraviolet rays
Answer:
The stratosphere
Explanation:
In the lower portion of the stratosphere there is a high concentration of ozone in relation to other parts of the atmosphere. This region is known as the ozone layer. This layer absorbs most of the Sun's ultraviolet radiation, especially ultraviolet B rays, allowing ultraviolet A rays to pass.
Answer:
TROPOSPHERE
Explanation:
Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate conduction between the pan and the burner?
Answer:
Earl needs to touch the pan to the burner.
Explanation:
Earl is demonstrating heat transfer by conduction.
Conduction is the transition of heat energy by direct contact. Hotter molecules have more momentum and they move faster.These hot molecules collide with the cooler ones, transfer their energy to cooler particles , this process is continued to pass it on to other molecule near to them.
Earl needs to touch the pan to the burner.
Answer:
D the answer is D
Explanation:
A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?
Answer:
v = 2 m/s (West)
Explanation:
Given
m = 2 Kg
v initial = 4 m/s (East)
F = 6 N (West)
t = 2 s
We can use the formula
v = v initial + a*t
if F = m*a ⇒ a = F/m = 6 N / 2 Kg
a = 3 m/s² (West)
then
v = v initial + a*t = (4 m/s) + (-3 m/s²)(2 s)
v = - 2 m/s = 2 m/s (West)
The velocity is a vector quantity. The velocity of the given cart is 2 m/s to the west.
To solve the given problem, use the kinematic equation,
[tex]v = s + at[/tex]........................1
Where,
[tex]s[/tex] - initial speed = 4 m/s
[tex]a[/tex] - acceleration
[tex]t[/tex] - time - 2 s
Acceleration can be calculated by the formula,
[tex]F = ma[/tex] or
[tex]a = \dfrac Fm[/tex]
Where,
m = mass = 2 Kg
F = force = 6 N
So,
[tex]a = \dfrac {6}2\\\\a = 3\rm \ m/s^2[/tex]
Put the values in equation 1,
[tex]v = 4 +3\times 2\\\\v = \bold {2 m/s}[/tex]
Therefore, the velocity of the given cart is 2 m/s.
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Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
Question:
Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
A typical asteroid in the asteroid belt, a trojan asteroid, a typical kuiper belt object, a typical oort cloud object
Answer:
Ranking of the objects from left to right based on their orbital period around the sun from shortest to longest
a typical asteroid in the asteroid belta Trojan asteroida typical Kuiper belta typical Oort cloud objectExplanation:
Half the mass of the belt is present in the largest four asteroids, they are ceres, vesta, pallas and hygiea. The asteroid belt total mass is 4% than that of moon, 22% Pluto. The three dwarf planets are present in the Kuiper belt Pluto, Haumea and Makemake. The spherical layer of icy objects around the sun are Oort cloud. Objects in the Oort cloud are made up of water ice, ammonia, and methane. It is spherical from outer and and torous shaped from inside.
The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant of variation.
Answer:
Explanation:
According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.
i.e. [tex]V\propto R[/tex]
[tex]V=R I[/tex]
where R=resistance
[tex]R\propto L[/tex]
[tex]R\propto \frac{1}{d^2}[/tex]
whee L and d are length and Diameter
thus [tex]R=k \frac{L}{d^2}[/tex]
where k=constant of Variation
Technician A says that the aspect ratio of a tire represents the relationship between the tire's cross-sectional height to its cross-sectional width. Technician B says that low aspect ratios provide a softer ride because they will deflect more over irregular surfaces and under heavy loads. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
Technician A
Explanation:
Often referred to as the profile or series, the aspect ratio of a tire is determined by dividing a tire’s section height by its section width when the tire is: inflated to maximum air pressure, mounted on the approved measuring rim, and under no load. This rules out Technician B.
A tire with a lower aspect ratio responds to lateral force more effectively than a tire with a higher aspect ratio. The aspect ratio affects steering stability. Generally, the shorter the sidewall, or the lower the aspect ratio, the less time it takes to transmit the steering input from the wheel to the tread. The result is quicker steering response. Aspect ratio also affects the tread contact patch. As a rule, a low-profile tire produces a wider tread contact patch. This wider tread contact patch creates a stiffer footprint that reduces distortion and provides improved cornering traction. Aspect ratio also impacts ride. A low-profile tire usually has a stiffer ride than the standard aspect ratio of 75 or more.
Final answer:
A. Technician A is correct about the aspect ratio of a tyre, while Technician B is incorrect about the impact of low aspect ratios on ride softness.
Explanation:
Technician A is correct. The aspect ratio of a tyre indeed represents the relationship between the tyre's cross-sectional height to its cross-sectional width. The aspect ratio is a critical factor in determining the tyre's performance and characteristics. Technician B is incorrect. Low aspect ratios do not provide a softer ride because they will deflect more over irregular surfaces.
In the given case, the aspect ratio primarily affects the tyre's handling, stability, and sidewall height. In conclusion, Technician A provides an accurate explanation regarding the aspect ratio of a tyre in relation to its dimensions and characteristics.
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizontal. To make his job easier, he oiled the bottom of the box, reducing the coefficient of friction to 0.250. What is the magnitude of the acceleration of the box?
Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the body is at rest and the pull on the chain is 6000 g (g is 9.8 m/s^2) The pull gets smaller uniformly at the rate of 360g N per each meter through which the body is raised. What is the velocity of the body when it has been raised 10m?
Answer:31.62 m/s
Explanation:
Given
mass of body [tex]m=400 kg[/tex]
Pull on chain is [tex]F_1=6000g N=60 kN[/tex]
Pull get smaller at the rate of [tex]F_2=360g N/m[/tex]
Net Upward Force [tex]F=6000 g-360 g\times 10=24 kN[/tex]
net acceleration [tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{24\times 1000}{m}[/tex]
[tex]a=\frac{24000}{400}[/tex]
[tex]a=60 m/s^2[/tex]
but g is acting downward
[tex]a_{net}=a-g=60-10=50 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
here initial velocity is zero
[tex]v^2=2\times 50\times 10[/tex]
[tex]v=31.62 m/s[/tex]
The velocity of the body will be "31.62 m/s".
Given:
Mass of body,
m = 400 kgPull on chain,
[tex]F_1= 6000 \ g.N[/tex][tex]F_2 = 360 \ g.N/m[/tex]The net upward force will be:
→ [tex]F = 6000-360\times 10[/tex]
[tex]= 6000-3600[/tex]
[tex]= 2400[/tex]
or,
[tex]= 24 \ kN[/tex]
Now,
The net acceleration will be:
→ [tex]a = \frac{F}{m}[/tex]
[tex]= \frac{24\times 1000}{400}[/tex]
[tex]= 60 \ m/s^2[/tex]
But,
"g" is acting downwards then,
→ [tex]a_{net} = a -g[/tex]
[tex]= 60-10[/tex]
[tex]= 50 \ m/s^2[/tex]
By using,
→ [tex]v^2-u^2=2as[/tex]
[tex]v^2= 2\times 50\times 10[/tex]
[tex]= 1000[/tex]
[tex]v = \sqrt{1000}[/tex]
[tex]= 31.62 \ m/s[/tex]
Thus the above answer is right.
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earth's troposphere hydrosphere and lithosphere contain large amounts of which element?
Answer: the element is Oxygen.
Explanation: Earth's troposphere hydrosphere and lithosphere contain large amounts of Oxygen. it is the most abundant element in the Earth's crust. Oxygen makes up 467,100 ppm (parts per million) of the Earth's crust, or 46.6%.
Oxygen is the primary element found in Earth's troposphere, hydrosphere, and lithosphere, as part of the air, water (H2O), and silicate materials in many minerals and rocks.
Explanation:The Earth's troposphere, hydrosphere, and lithosphere all contain large amounts of the element oxygen. The troposphere, part of the atmosphere, contains oxygen in the air we breathe, while the hydrosphere, which encompasses all water on Earth, would have oxygen as a component of H2O. In the lithosphere, which covers the crust and the upper part of the mantle, oxygen is found in many minerals and rocks as part of silicate materials.
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When following a motorcycle, drivers must remember that motorcycles can ____________ than other vehicles.
Answer:
Stooping more Quickly
Explanation: Stooping more quickly means that Motor cycles can easily bend and stop,this is due to it's design, Motor cycle have only two wheels,they contain just two tyres one in front and one at the back,they are not as firm on the ground when compared to other vehicles such Motor car. They have a lower weight which makes them easily bent or stopped once there's any barrier.
Motorcycles can stop quicker than other vehicles.
Explanation:When following a motorcycle, drivers must remember that motorcycles can stop quicker than other vehicles. Due to their smaller size and weight, motorcycles have a shorter stopping distance. This means that drivers should maintain a safe following distance to avoid rear-ending the motorcycle.
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Which element has properties of electrical conductivity and luster and exists as a liquid at STP?
Answer:
Mercury - hydrargyrum (Hg)
Explanation:Mercury is a chemical element with the symbol Hg and atomic number 80. It is commonly known as quicksilver and was formerly named hydrargyrum.
Symbol: Hg
Atomic number: 80
Atomic mass: 200.59 u
Melting point: -38.83 °C
Boiling point: 356.7 °C
Electron configuration: [Xe] 4d^14 5d^10 6s^2
Group :12
Ionization energy 1007.1J
Electronegativity :2
Crust Abundance : 0.085ppm
Electron Affinity : Unknown eV
Name: Mercury , the first planet in the solar system (Hg from former name hydrargyrum , from greek hydr - water and argyros silver .
I hope it helps. :).
A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. All of the following are true EXCEPT:_________
I. the angle of reflection is 30 degrees
II. the angle of refraction is 30 degrees
III. total internal reflection will result, depending on the critical angle
Answer:
All of the following are true EXCEPT, the angle of refraction is 30 degrees
Explanation:
It is given that,
The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. We know that when a ray of white light moves through the air and strikes the surface of water in a beaker, refraction occurs. Let r is the angle of refraction. It can be calculated using Snell's law as :
[tex]n=\dfrac{sin\ i}{sin\ r}[/tex]
[tex]sin\ r=\dfrac{sin\ i}{n}[/tex]
[tex]sin\ r=\dfrac{sin(30)}{1.33}[/tex]
[tex]r=22.08^{\circ}[/tex]
So, the angle of refraction is 22.08 degrees.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since, it is incident at an angle of 30 degrees, so the angle of reflection will be 30 degrees. Also, total internal reflection will result, depending on the critical angle.
So, the wrong statement is the angle of refraction is 30 degrees. It is equal to 22.08 degrees.
why is 21-cm radiation so important to the study of interstellar matter and the galaxy
Answer:
Because its the most suitable radiation for that use
Explanation:
Astronomers use the 21 cm radiation for interstellar and cosmological research as this type of radiation is best suitable for the purpose its intended for. It can easily go through gas and dust molecules and other interstellar bodies that causes a blur vision of the content of space.
Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface?
The collision is elastic.
A. All of the initial kinetic energy of the less-massive object is lost.
B. The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less massive object.
C. The speed of the objects that are stuck together will be less than the initial speed of the less-massive object.
D. The direction of motion of the objects that are stuck together depends on whether the hit is a head on collision.
Answer:
C. The initial momentum should be equal to the final momentum due to the conservation of momentum.
[tex]P_{initial} = mv_0\\P_{final} = (M+m)v_1\\v_1 = \frac{m}{M+m}v_0[/tex]
Since m/(M+m) < 1, v_1 > v_0.
Explanation:
Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.
Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.
Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.
Answer:
[tex]\Delta l=0.015m[/tex]
Explanation:
We have given initial length of the steel guitar l = 1 m
Cross sectional area [tex]A=0.5mm^2=0.5\times 10^{-6}m^2[/tex]
Young's modulus [tex]\gamma=2\times 10^{11}Pa[/tex]
Force F = 1500 N
So stress [tex]=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa[/tex]
We know that young's modulus [tex]=\frac{stress}{strain}[/tex]
So [tex]2\times 10^{11}=\frac{3\times 10^{9}}{strain}[/tex]
[tex]strain=1.5\times 10^{-2}=0.015m[/tex]
Now strain [tex]=\frac{\Delta l}{l}[/tex]
[tex]0.015=\frac{\Delta l}{1}[/tex]
[tex]\Delta l=0.015m[/tex]
Answer:
Explanation:
L = 1 m
A = 0.5 mm² = 0.5 x 10^-6 m²
Y = 2 x 10^11 Pa
F = 1500 N
ΔL = ?
Use the formula for the young's modulus
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]\Delta L = \frac{FL}{AY}[/tex]
[tex]\Delta L = \frac{1500\times 1}{0.5\times10^{-6}\times 2\times 10^{11}}[/tex]
ΔL = 0.015 m
ΔL = 0.02 m
Suppose you have an accounting background and you are considered an expert in the field of discovering errors in electronic spreadsheets. You are considered a knowledge engineer if you are part of a team that is developing an expert system to discover errors in spreadsheets. True or false?
Answer: True
Explanation: A knowledge engineer is a professional who works to improve and builds with the aid of science and also implements advanced logic and AI( artificial intelligence) in computer systems in order to investigate particular issues.
A knowledge engineer also helps tries to emulate the judgements of man.
Since you are part of the team involved in discovering errors in the electronic spread sheet you are a knowledge engineer.
Technician A says that the connecting rods should be marked before disassembly. Technician B says that pistons are cam ground so that when operating temperature is reached, the piston will have expanded enough across the piston pin area to become oval shaped. Who is right?
A) Technician A only B) Technician B only
C) Both technicians A and B D) Neither technician A nor B
Answer:
C.
Explanation:
Both technicians A & B are right since the connecting rods must be marked to indicate which surface faces front. And a cam-ground piston machine must also be configured oval shape so that it will be able to fit the cylinder better throughout its operating temperature range.
Technician A is correct that the connecting rods should be marked before disassembly. Technician B is incorrect about the cam ground pistons. The correct answer is A) Technician A only.
Explanation:Technician A is correct that the connecting rods should be marked before disassembly. This is important in order to ensure that the rods are reinstalled in the same location and orientation as before, which helps maintain balance and proper engine function.
Technician B is incorrect. Pistons are not cam ground to become oval shaped when operating temperature is reached. Instead, pistons are typically designed with expansion clearances to account for thermal expansion, ensuring that they maintain proper shape and function at operating temperatures.
Therefore, the correct answer is A) Technician A only.
Which two security measures must an engineer follow when implementing layer 2 and layer 3 network design?
Answer:
Utilize the Private VLAN feature to segregate network traffic at Layer 2, and
Utilize the ARP inspection feature to help prevent the misuse of gARP.
Explanation:
An Engineer designing layer 2 and layer 3 network, must Utilize the Private VLAN feature to segregate network traffic at Layer 2 and Utilize the ARP inspection feature to help prevent the misuse of gARP.
Consider a telescope with a small circular aperture of diameter 2.0 centimeters.A) If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resolved if their angular separation is 3.0×10−5 radians?
Answer:
[tex]4.91803\times 10^{-7}\ m[/tex]
Explanation:
[tex]\theta[/tex] = Angular seperation = [tex]3\times 10^{-5}\ rad[/tex]
[tex]\lambda[/tex] = Wavelength
We have relation
[tex]sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow \lambda=\dfrac{dsin\theta}{1.22}\\\Rightarrow \lambda=\dfrac{2\times 10^{-2}sin(3\times 10^{-5})}{1.22}\\\Rightarrow \lambda=4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength of the telescope is [tex]4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
Given data:
The diameter of the telescope is, d = 2 cm = 0.02 m.
The angular separation of the two point source of light is, [tex]\theta = 3.0 \times 10^{-5} \;\rm radians = 3.0 \times 10^{-5} \;\rm radians \times \dfrac{\pi}{180}\\\theta = 5.23 \times10^{-7}^{\circ}[/tex].
The angular separation of two point source is also known as apparent distance between the sources, and it is defined as the angle between the two objects as viewed directly by the observer.
The standard relation for the angular separation of telescope is given as,
[tex]sin \theta = 1.22 \times \dfrac{\lambda}{d}[/tex]
here, [tex]\lambda[/tex] is the maximum wavelength of light.
Solving as,
[tex]sin (5.23 \times 10^{-7}) = 1.22 \times \dfrac{\lambda}{0.02}\\\\\lambda = \dfrac{sin (5.23 \times 10^{-7}) \times 0.02}{1.22} \\\\\lambda = 1.49 \times 10^{-10} \;\rm m[/tex]
Thus, we can conclude that the maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
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What is the difference between balanced and unbalanced forces
Balanced forces are equal and opposite, causing no change in an object's movement, leading to static equilibrium. On the contrary, unbalanced forces change an object's velocity or direction, causing it to move or accelerate.
Explanation:The primary difference between balanced and unbalanced forces lies in their effects on the motion of an object. Balanced forces are equal and opposite forces that act on an object, resulting in no change in its velocity or direction. An example would be a static car or an ice hockey stick lying flat with two equal and opposite forces applied to it. Friction is negligible and the gravitational force is balanced by the support, leading to static equilibrium.
On the other hand, unbalanced forces are not equaled.
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Tarzan steps out of his house and grabs a hanging vine to swing to the ground. If his house is 11.3 m above the ground, what is Tarzan's speed when he reaches the ground?
Answer:
v = 14.88 m / s
Explanation:
For this exercise let's use energy conservation. We place the reference system on the ground
Start. Tarzan before jumping
Em₀ = U = m g h
Final. Upon reaching the ground
[tex]Em_{f}[/tex]= K = ½ m v²
Energy is conserved
Em₀ = [tex]Em_{f}[/tex]
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 11.3)
v = 14.88 m / s
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
Answer: the correct option is B ( on the half of the road that the vehicle is traveling).
Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is
on the half of the road that the vehicle is traveling.
Answer:
B
Explanation:
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
According to the law regrading pedestrian, the pedestrian has the right of way immediately she is crossing on the crosswalk and is half of the road that the vehicle is travelling.
These rules are there for the safety of the driver, pedestrians and other road users
The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion?
Question:
The theory of force generation proposed is best supported by which of the following observations about Amoeba locomotion?
a. amoeboid movement stops upon expose to cytochalasins
b. amoeboid movement cannot occur if mitosis is blocked
c. moving amoeba cells produce more troponin than do stationary ones
d. the rate of movement is inversely proportional to the viscosity of the medium in the Amoeba moves
Answer:
The amoeba location can be supported by :The amoeboid movement stops upon expose to cytochalasins
Explanation:
The amoeboid movement common in eukaryotic cells. cytochalasins are the type of drug that inhibits the growth of the micro filaments, micro filaments capability to elongate are both implemented as important to make the force for movement in amoeba. further Cytochalasins are known as the fungal metabolites which can bind the actin filaments and also restrict polymerization and elongation of actin. It can also permeate cell membranes, preventing the cellular translocation and cause cells to enucleate.
A particular star cluster contains stars all with the same apparent magnitude of +4. Near the cluster (and at the same distance from the Earth) is a single star with an apparent magnitude of +1. The star's brightness appears to be the same as the collective brightness of the entire cluster.
How many stars are in this cluster?
Answer:
3,000 stars.
Explanation:
Subtract the magnitude of the single star from the magnitude of the star cluster.
(+4) - (+1) = +3
Then multiply your answer by 1000
= 3,000 is a stars.
What types of compounds are involved in solutions that are used to grow crystals? In your own words please
Answer:
This is a table of common chemicals that produce nice crystals. The color and shape of the crystals are included. Many of these chemicals are available in your home. Other chemicals in this list are readily available online and are safe enough for growing crystals at home or in a school. Recipes and specific instructions are available for hyperlinked chemicals.
Ionic compounds, held together by electrostatic forces, form crystalline structures that exhibit rigid and brittle properties. They tend to have high melting and boiling points, reflecting the strength of the ionic bonds. Though ionic solids are poor conductors in their solid state, they readily dissolve in water, becoming good conductors when melted or dissolved.
Explanation:Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
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Why are summer days longer than winter days on earth?
Answer:
Tilt of the Earth
Explanation:
The Earth's rotational axis is titled at an angle 23.4°. When then northern hemisphere is pointed towards the Sun it is summer in the northern hemisphere and winter in the southern hemisphere.
In the northern hemisphere a day would be longer and gets more sunlight than the southern hemisphere.
The same happens when it is summer in the southern hemisphere.
Answer:
The sun is father away during the winter.
Explanation:
In the northern hemisphere, during the winter it is summer in the southern hemisphere. The sun is farther away for us so the days get shorter in the winter since the sun inches away faster.
Let v⃗ A be the velocity of the car at point A. What can you say about the acceleration of the car at that point?
Answer and Explanation:
Let the velocity of the car at a point is A is v
We have to tell about the acceleration at that point
Acceleration at that point will be perpendicular to the velocity and directed inside of the track
We can tell anything about the magnitude of the acceleration as magnitude of acceleration is the rate of change of velocity and here we have no information about time
While a car's velocity at a given point only describes its speed and direction at that moment, its acceleration at that point accounts for changes in speed and direction over time. To ascertain acceleration at a point, the velocities before and after that point need to be compared.
Explanation:The velocity of a car at a given point, such as point A, tells us it's direction and speed at that instant. However, the acceleration of the car at that point is a different matter. Acceleration is defined as the change in velocity per unit time, and it doesn't only involve speed, but also direction. A car could be moving at a constant speed, but if it's changing direction, it's considered to be accelerating. Therefore, to ascertain acceleration at point A, we would need to know not just the velocity at point A, but also at points immediately before and after.
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The total lifetime of the Sun is about 10 billion years, from when it was born to when it dies, and the half-life of 238U is 4.5 billion years. Suppose a particular meteoroid contained 24 micrograms of uranium-238 (238U) when the Sun was born. If the meteoroid is still in orbit when the Sun dies, it will contain _____
Answer:
a little less than 6 micrograms
Explanation:
The half-life of an element is defined as the time in which half of the isotope emits its radiation and becomes a different element. Therefore after 4.5 billion years the meteoroid will contain 12 micrograms of uranium, after 9 billion years will contain 6 micrograms. After 10 billion years will contain a little less than 6 micrograms.