Answer:
[tex]d=490\ m[/tex] is his final displacement from the point A after 60 seconds.
Explanation:
Given:
Cyclist is moving away from A.
velocity of cyclist, [tex]v=4.1\ m.s^{-1}[/tex]displacement of the cyclist from point A at the time of observation, [tex]d_i=244\ m[/tex]time after which the next observation is to be recorded, [tex]t=60\ s[/tex]Now as the cyclist is moving away from point A his change in displacement after the mentioned time:
[tex]\Delta d=v.t[/tex]
[tex]\Delta d = 4.1 \times 60[/tex]
[tex]\Delta d=246\ m[/tex]
Now the the final displacement from point A after the mentioned time:
[tex]d=d_i+\Delta d[/tex]
[tex]d=244+246[/tex]
[tex]d=490\ m[/tex]
A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it
Answer:
The work done by the hoop is equal to 5.529 Joules.
Explanation:
Given that,
Mass of the hoop, m = 96 kg
The speed of the center of mass, v = 0.24 m/s
To find,
The work done by the hoop.
Solution,
The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,
[tex]K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
I is the moment of inertia, [tex]I=mr^2[/tex]
Since, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]K_i=mv^2[/tex]
[tex]K_i=96\times (0.24)^2=5.529\ J[/tex]
Finally it stops, so the final energy of the hoop will be, [tex]K_f=0[/tex]
The work done by the hoop is equal to the change in kinetic energy as :
[tex]W=K_f-K_i[/tex]
W = -5.529 Joules
So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.
Wo baseballs are fired into a pile of hay. If one has twice the speed of the other, how much farther does the faster baseball penetrate?
Answer:
Explanation:
Given
Two baseballs are fired into a pile of hay such that one has twice the speed of the other.
suppose u is the velocity of first baseball
so velocity of second ball is 2u
suppose [tex]d_1[/tex] and [tex]d_2[/tex] are the penetration by first and second ball
using [tex]v^2-u^2=2 ad[/tex]
where v=final velocity
u=initial velocity
a=acceleration
d=displacement
here v=0 because ball finally stops
[tex]0-u^2=2ad_1----1[/tex]
for second ball
[tex]0-(2u)^2=2ad_2----2[/tex]
divide 1 and 2 we get
[tex]\frac{u^2}{4u^2}=\frac{d_1}{d_2}[/tex]
as deceleration provided by pile will be same
[tex]\frac{1}{4}=\frac{d_1}{d_2}[/tex]
[tex]d_2=4d_1[/tex]
thus faster ball penetrates 4 times of first ball
Final answer:
The faster baseball will penetrate farther into the pile of hay due to the greater change in kinetic energy caused by its higher speed.
Explanation:
In this scenario, we can analyze the problem using the concept of work and energy. When the slower baseball and the faster baseball are both fired into the pile of hay, the work done by air resistance on each baseball will be different. The work done by air resistance is equal to the change in kinetic energy of the baseball. Since the faster baseball has twice the speed of the slower baseball, it will experience a greater change in kinetic energy and therefore penetrate farther into the pile of hay.
What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above horizontal?
Answer:
Length of shadow cast by hand= 5.0 cm
Explanation:
With the hand inclined at an angle θ=30.0∘ above horizontal, it can be imagined to form part of a right angled triangle that has the following parts
vertical screen=vertical part of the triangle where we cast the shadow of the hand
The hand=the hypotenuse side
The horizontal = side to which the hand makes the 30.0° angle
By trigonometric relationship
Sin θ°= Opposite/hypotenuse
Sin θ° = (Shadow cast by hand)/(Length of hand)
Sin 30° = ( length of shadow cast by hand)/10cm
or 0.5 = (length of shadow cast by hand)/10cm
Length of shadow cast by hand= 0.5 × 10cm = 5.0 cm
When transiting a great distance a Navigator prepares a __________ track so the ship can steady courses while driving the shortest distance.
Answer:
composite track
Explanation:
To travel an incredible circular track, the guide must constantly change course because the extraordinary circular track is a turn when plotted on a Mercator map. It is ridiculous to try to navigate an incredible circular route. All things considered, to make the best use of the shorter cruise separation from the extraordinary circular runway, pilots generally divide an incredible hover runway between the underlying position and the target into many much smaller sections ( for trajectory purposes) of approximately one to several days of cruising time (based clearly on the specialty and conditions) and making course changes every day simultaneously, generally in the early afternoon. Absolute separation is thus the set of separations of these fragments determined by the methods of Mercator Sailing. A potential problem with the incredible circular track, however, is the most limited route between two areas, similarly for most tracks closer to the well (or at a higher range) than the two points, starting point or goal. The high areas are often in danger due to the terrible climate and icing. A protected thought of a veteran sailor is to set a range limit for the long voyage plan. This arrangement is called an extraordinary composite circle course arrangement, terminated with way points. This minicomputer soothes the monotonous procedure for deciding these way points for travel.
Answer:
The answer is composite track.
Explanation:
For ships to cover the shortest distance between two points on the surface of the earth, navigators base their calculations using great circles. A great circle is circular line drawn on a globe that follows the circumference of the earth ( thereby dividing the globe into equal halves ). As the ship moves from one point to the other, the navigator adjusts its course because the earth is on a constant rotation. Great circles, because they usually cover distances of about 40,000km are broken down into smaller lines called Rhumb so as to provide a steady course. The one common great circle is the equator and the ship's heading does not change on this line.
In the equation: v3 *t/A where v is the speed, t is the time and A is the area. Which of the following represents the dimensions of the result. Select one: a. length*time b. length2 /time c. length/time2 d. 1/time
Answer:
option C
Explanation:
given,
[tex] \dfrac{v^3t}{A}[/tex]
where v is the speed
t is the time
A is the area
we know v = m/s
dimension of v = [LT⁻¹]
v³ = [L³ T⁻³]
t = s
dimension of t = [T]
unit of area = m²
dimension of A = [L²]
dimension of the given equation
[tex]\dfrac{v^3t}{A}= \dfrac{[L^3T^{-3}][T]}{[L^2]}[/tex]
[tex]\dfrac{v^3t}{A}= \dfrac{[L]}{[T^2]}[/tex]
=[tex]\dfrac{Length}{time^2}[/tex]
hence, the correct answer is option C
Final answer:
In the equation [tex]v^3[/tex] * t / A, where v is speed, t is time, and A is the area, the result has the dimensions of length/time squared (L/[tex]T^2[/tex]), which represents an acceleration. The correct answer is c. [tex]length/time^2[/tex].
Explanation:
The equation given is [tex]v^3[/tex] * t / A. To determine the dimensions of the result, we need to assess each variable's dimensional formula. The velocity (v) has the dimension of length/time (L/T). Time (t) is already in the dimension of time (T). The area (A) has the dimension of length squared ([tex]L^2[/tex]).
Combining these, we have the dimensions for velocity cubed as [tex](L/T)^3[/tex], which gives us [tex]L^3/T^3[/tex]. This is multiplied by the time (T), resulting in [tex]L^3/T^2[/tex]. When we divide this by the area ([tex]L^2[/tex]), the result simplifies to L/T, which is the dimension of velocity. However, because we are dividing by area and not velocity, the final dimension is L/[tex]T^2[/tex], which is option c: [tex]length/time^2[/tex].
An unknown solid with a mass of 2.00 kilograms remains in the solid state while it absorbs 32.0 kilojoules of heat. Its temperature rises 4.00 degrees Celsius. What is the specific heat of the unknown solid?
Answer: The specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
Explanation:
As we know that,
[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex] (1)
where,
q = heat absorbed = 32.0 kJ = [tex]32.0\times 10^3J[/tex] J (1kg=1000g)
[tex]m[/tex] = mass of unknown solid= 2.00 kg = [tex]2.00\times 10^3g[/tex] (1kg=1000g)
[tex]T_{final}[/tex] = final temperature
[tex]T_{initial}[/tex] = initial temperature
[tex]\Delta T[/tex] =[tex]4.00^0C[/tex]
[tex]c[/tex] = specific heat of unknown solid = ?
Now put all the given values in equation (1), we get
[tex]32.0\times 10^3J=2.00\times 10^3g\times c\times (4.00^0C)][/tex]
[tex]c=4.00J/g^0C[/tex]
Therefore, the specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
The scientific heat of the unknown solid will be "4.00 J/g°C".
Specific heat:Given values are:
Heat absorbed, q = 32.0 kJ or, [tex]32.0\times 10^3[/tex] J
Mass, m = 2.00 kg or, [tex]2.00\times 10^3[/tex] g
Rise in temperature, ΔT = 4.00°C
We know the relation,
→ q = m×c×ΔT
or,
→ = m×c×([tex]T_{final} - T_{initial}[/tex])
By substituting the values,
[tex]32.0\times 10^3=2.00\times 10^3\times c\times 4.00[/tex]
[tex]c = 4.00[/tex] J/g°C
Thus the above answer is appropriate.
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Dr. Wolski does research on the potential relationship between neurotransmitter deficiencies and mood states. Which psychological specialty does Dr. Wolski's research best represent
Answer:
Bio psychology
Explanation:
Since, Dr. Wolski does research on the potential relationship between neurotransmitter deficiencies and mood states. His psychological specialty must be biological psychology.
Biological psychology, or bio psychology, is a discipline in which scientific investigation and clinical practice investigates the relation between brain and body. In this area scholars evaluate the biological basis of feelings, impulses and behaviors.
An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remainder of the race. Find the runner's speed (in m/s) at the following times.
Answer:
The runner's speed at the following times would remain 8.64 m/s.
Explanation:
Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.
In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.
First equation of motion: Vf = Vi + a×t
Vf stands for final velocity
Vi stands for initial velocity
a stands for acceleration
t stands for time
In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.
The acceleration (a) is given as 2.4 m/s²
The time (t) is given as 3.6 s
Now put the values of Vi, a and t in first equation of motion
Vf = Vi + a×t
Vf = 0 + 2.4×3.6
Vf = 2.4×3.6
Vf = 8.64 m/s
So,the runner's speed at the following times would remain 8.64 m/s.
Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint of the distance is marked to make it easier to see how the locations compare. Assume the spaceship has the same mass throughout the trip (that is, it is not burning any fuel). Rank the five positions of the spaceship from left to right based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest.
The five positions of the spaceship from left to right are based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest is [tex]5, 1, 2, 4, 3[/tex]
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another.
The gravitation force that Earth exerts on the spaceship will be:[tex]F_{ES}=(Gm_1m_E)/r^2[/tex]
Where [tex]F_{ES}[/tex] the force exerted on the spaceship by Earth [tex]m_1\\\\[/tex] is the mass of the spaceship and r is the distance between the.
[tex]F_{ES}\ \alpha\ 1/r^2[/tex]
This indicates larger the distance smaller will be the force. The correct order is [tex]5, 1, 2, 4, 3[/tex].
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The strength of the gravitational force that Earth exerts on a spaceship varies depending on the distance between them. The force is strongest when the spaceship is closest to Earth and weakest when it is closest to the Moon.
Explanation:Position 1: The spaceship is closest to Earth, so the gravitational force is strongest here. Position 2: The spaceship is moving away from Earth, so the gravitational force is slightly weaker than at Position 1 but stronger than at the other positions. Position 3: The spaceship is at the midpoint between Earth and the Moon, so the gravitational force is weaker than at Positions 1 and 2 but still stronger than at Positions 4 and 5. Position 4: The spaceship is closer to the Moon than to Earth, so the gravitational force from the Moon is stronger than the force from Earth. Position 5: The spaceship is closest to the Moon, so the gravitational force from the Moon is strongest here, and the force from Earth is weakest. Learn more about Gravitational force here:
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At standard temperature and pressure, a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas have the same A. average molecular knetic energyB. average molecular speedC. volumeD. effusion rateE. density
Answer:
The correct answer is option A.
Explanation:
The average kinetic energy of the gas particle only depends upon the temperature of the gas.
The formula for average kinetic energy is:
[tex]K.E=\frac{3}{2}kT[/tex]
where,
k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]
T = temperature
So, at standard temperature and pressure 0.50 moles of hydrogen and 1.0 mole of oxygen sample will have same value of average kinetic energy.
Where as in other option enlisted in question , molar masses of both gases will be involved which will give different answers for both the gases.
This question involves the concepts of the average kinetic energy of the molecules of a gas and temperature.
The gases will have the same "A. average molecular kinetic energy".
The average kinetic energy of gas molecules is given by the following formula:
[tex]K.E = \frac{3}{2}KT[/tex]
where,
K.E = average kinetic energy
K = Boltzmann constant
T = absolute temperature
Hence, the average kinetic energy depends upon the absolute temperature only. Since the temperature is the same for both gases. Hence, their average kinetic energy will also be the same.
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Your car's engine is a heat engine; it converts the thermal energy from burning fuel into energy to move your car and power its systems. On a cold winter day, you needn't feel guilty about cranking up the heat in your car; running the heater doesn't cost any additional energy beyond the small amount needed to run the fan. Explain why this is so.
Answer:
Explanation:
The efficiency of the heat engine is less than 100%. The heat engine converts the thermal energy from the burning of fuel to the mechanival energy. Thus, some of the energy gets lost as waste heat.
Hence, this waste heat is used up to heat up the interior of the car and person feels warmer in the car.
Thus, the waste heat generated in the engine is used for the heating of the car interior.
Final answer:
Running the heater in a car on a cold winter day doesn't cost much additional energy beyond running the fan because the heat generated by the engine is mostly released into the environment instead of being used for work in the car.
Explanation:
Heat engines convert thermal energy into work. In a car engine, heat is produced when fuel is burned, and this heat is converted into kinetic energy to move the car and power its systems. Running the heater in a car on a cold winter day doesn't cost much additional energy beyond running the fan because the heat generated by the engine is mostly released into the environment instead of being used for work in the car.
The brakes on your automobile are capable of creating a deceleration of 4.9 m/s2. If you are going 149 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 100 km/h speed limit?
Answer:
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
Explanation:
Hi there!
The equation of velocity for an object moving in a straight line is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time.
We have to find the time at which the velocity is 100 km/h with a decceleration of 4.9 m/s² and an initial velocity of 149 km/h. Let´s first convert km/h into m/s:
149 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 41.4 m/s
100 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 27.8 m/s
Now, let´s solve the equation of velocity for the time:
v = v0 + a · t
(v - v0) / a = t
Replacing with the data:
(27.8 m/s - 41.4 m/s) / -4.9 m/s² = t
Notice that the acceleration is negative because you are slowing down.
t = 2.8 s
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
To get the car under the speed limit, it would take approximately 4.34 seconds.
Explanation:To calculate the minimum time it takes for the car to get under the 100 km/h speed limit, we need to find the deceleration required to slow down from 149 km/h to 100 km/h.
First, let's convert the speeds to m/s. 149 km/h is equal to 41.4 m/s and 100 km/h is equal to 27.8 m/s.
Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange the equation to solve for acceleration:
a = (v² - u²) / (2s)
Plugging in the values, we get a = (27.8² - 41.4²) / (2(-30.5)).
Solving this equation, we find that the minimum deceleration required is approximately -3.13 m/s².
Finally, we can use the formula a = Δv / t to find the minimum time:
t = Δv / a = (41.4 - 27.8) / 3.13 = 4.34 seconds.
Which layer in the Earth has a composition similar to the granite shown in this photograph?
Answer:
Earth crust and specifically the continental crust.
Explanation:
If we examine the earth crust there is mostly the granite and basalt and most of the granite is present in the continental crust part which is less thicker and denser. That's why we say that the continental crust has the composition similar to that of granite.
A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of the ball's momentum is 0.800 kg · m/s just before it lands on the ground?
Answer:
1.55 m
Explanation:
Momentum: This can be defined as the product of mass of a body and it velocity. the S.I unit of momentum is kgm/s.
Mathematically,
Momentum can be represented as,
M = mv................................. Equation 1
Where m = mass of the body, v = velocity of the body, M = momentum.
Making v the subject of the equation,
v = M/m........................................... Equation 2
Given: M = 0.80 kg.m/s, m = 0.145 kg.
Substituting into equation 2,
v = 0.8/0.145
v = 5.52 m/s.
Using the equation of motion,
v² = u² + 2gs ....................... Equation 3.
Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.
Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².
Substituting into equation 2
5.52² = 0² + 2(9.81)s
30.47 = 19.62s
s = 30.47/19.62
s = 1.55 m.
Thus the ball was dropped from a height of 1.55 m
A person has been exposed to a particular antigen and now experiences a repeat exposure. What stimulates a quicker immune response?a) Memory T cellsb) immunityc) antibodiesd) macrophages
Answer:
a. Memory T cells
Explanation:
Memory T cells are actually the antigen-specific T cells that remain long-term after an infection has been eliminated. These memory T cells are quickly converted into large numbers of effector T cells upon reexposure to the specific invading antigen, thus providing a rapid response to past infection that has been experienced before
_____ is a disorder that results from damage to the brain's motor centers, causing difficulty with motor control so that speech and movements are impaired. Klinefelter syndrome Muscular dystrophy Cerebral palsy Crohn's disease.
Answer:
Cerebral palsy
Explanation:
Cerebral palsy - it is referred to that disorder which is related to damages that caused permanent disorder in the functioning of body parts. it affects the proper functioning of muscles thus cause the coordination problem.
it is caused due to abnormalities in the brain that result in the coordination of the body. As it is related to abnormalities in the brain thus it also causes a problem in vision, speaking, hearing, etc
In the manufacturing of computer chips, cylinders of silicon are cut into thin wafers that are 3.00 in. in diameter and have a mass of 1.50 g of silicon. How thick, in millimeters, is each wafer if silicon has a density of?
The thickness of a silicon wafer can be found using the volume of a cylinder and the density formula. Given the diameter and the mass, rearranging to solve for height will give the thickness of the wafer in millimeters.
Explanation:The thickness of the silicon wafer can be found by using the formula for the volume of a cylinder (Volume = pi * radius2 * height) and the definition of density (Density = mass/volume). Given the silicon wafer has a diameter of 3 inches (or a radius of 1.5 inches), and a mass of 1.5g, we can determine the volume of the wafer from the given density. Rearranging the equation for the volume of a cylinder to solve for height (or thickness, in this case) gives: Thickness = Volume / (pi * radius2). Assuming measurements are converted correctly for consistent units, this calculation will give the thickness of the wafer in millimeters.
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The force exerted on the small piston of a hydraulic lift is 780 N . If the area of the small piston is 0.0075 m2 and the area of the large piston is 0.13 m2, what is the force exerted by the large piston?
Answer:
13520 N
Explanation:
Pascal's Principle: The principle states that the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
The operation of hydraulic press and car brake system is based on pascal's principle.
From pascal's principle,
F/A = f/a ........................... Equation 1
Where F = force exerted by the large piston, A = area of the large piston, f = force applied to the small piston, area of the small piston.
Making F the subject of the equation
F = A(f/a)......................... Equation 2
Given: A = 0.13 m², a = 0.0075 m², f = 780 N
Substituting into equation 2
F = 0.13(780/0.0075)
F = 13520 N.
Thus the force exerted by the large piston = 13520 N
A 12-kg lead brick falls from a height of 1.9 m. (c) The brick falls onto a carpet, 2.0 cm thick. Assuming the force stopping it is constant, find the average force the carpet exerts on the brick.
Answer:
F = -11199.63 N
Explanation:
given,
mass of the brick = 12 Kg
height of the fall, h = 1.9 m
thickness of the carpet = 2 cm = 0.02 m
average force = ?
velocity of brick just before hitting mat
[tex] v = \sqrt{2gh}[/tex]
[tex]v =\sqrt{2\times 9.81\times 1.9}[/tex]
v = 6.11 m/s
velocity of brick just before hitting ground= 6.11 m/s
final velocity = 0 m/s
using equation of motion for acceleration calculation.
v² = u² + 2 a s
0² = 6.11² + 2x a x 0.02
[tex]a = -\dfrac{6.11^2}{0.04}[/tex]
a =-933.3025 m/s²
now, average force is equal to
F = m a
F = 12 x (-933.3025)
F = -11199.63 N
negative sign represent the decelerating force.
What is the difference between a continuous spectrum and a line spectrum
Answer:
Explanation:
The continuous spectrum is a band of visible colors of light. The continuous spectrum contains all the colors of all the visible wavelengths. Usually, most of the light is emitted from a single source. Whereas a line spectrum contains only a few colors and wavelengths of the visible spectrum with gaps of the discontinuity between them. The line spectrum is usually emitted by an excited electron of an atom that is going back to its ground state.A continuous spectrum shows all colors of the rainbow with no gaps, produced by a solid or very dense gas emitting radiation. In contrast, a line spectrum consists of only certain discrete wavelengths - either as an absorption spectrum with dark lines representing absorbed wavelengths, or as an emission spectrum showing bright lines for emitted wavelengths from excited gas atoms.
Explanation:The difference between a continuous spectrum and a line spectrum mainly lies in the type of light they represent. A continuous spectrum is formed when a solid or a very dense gas gives off radiation, showing an array of all wavelengths or colors of the rainbow. This can be seen when white light is passed through a prism as represented in Figure 5.10. It's like viewing a rainbow where all the colours blend into each other without any gaps.
On the other hand, a line spectrum, which could either be an absorption or an emission spectrum, consists of light in which only certain discrete wavelengths are present. Absorption spectrum appears as a pattern of dark lines or missing colors superimposed on the continuous spectrum, created when a cloud of gas absorbs certain wavelengths from the continuous spectrum behind it. Meanwhile, an emission spectrum appears as a series of bright lines when we examine an excited gas cloud, demonstrating that the gas is emitting light at only certain wavelengths, as showcased in Figure 5.12.
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A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the atmospheric pressure is 100 kPa, determine the final temperature of the gas in °C.
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
[tex]\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\[/tex]
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
The problem involves using Charles's Law, a form of the ideal gas law, to find the final temperature of nitrogen after heating, by converting all pressures to absolute pressures and applying the law to relate initial and final states.
Explanation:The student's question involves determining the final temperature of nitrogen gas in a rigid tank after heating, given initial temperature and pressures, using the ideal gas law. To solve this problem, we assume the nitrogen behaves as an ideal gas and use the relation between pressure, volume, temperature, and the number of moles of gas, which is constant since the tank is rigid.
Firstly, convert all pressures into absolute pressures by adding atmospheric pressure to the gage pressures. Then, apply the ideal gas law in the form of Charles's Law (P1/T1 = P2/T2), which relates pressure and temperature at constant volume and number of moles. To find the final temperature (T2), rearrange the equation to T2 = (P2/P1) * T1, where P2 and P1 are the final and initial absolute pressures, respectively, and T1 is the initial temperature in Kelvin.
When 1.60 × 10 5 J 1.60×105 J of heat transfer occurs into a meat pie initially at 17.5 °C , 17.5 °C, its entropy increases by 485 J / K . 485 J/K. Estimate the final temperature of the pie.
Answer:
Explanation:
Given
Heat transfer [tex]Q=1.6\times 10^5\ J[/tex]
initial Temperature [tex]T_i=17.5^{\circ}\approx 290.5\ K[/tex]
Entropy change [tex]dS=485\ J/K[/tex]
The expression for entropy is given by
[tex]dQ=TdS[/tex]
[tex]T=\frac{dQ}{dS}[/tex]
[tex]T=\frac{1.6\times 10^5}{485}[/tex]
[tex]T=329.89\ K[/tex]
Temperature can be written as average of initial and final temperature
[tex]T=\frac{T_i+T_f}{2}[/tex]
[tex]329.89=\frac{T_f+290.5}{2}[/tex]
[tex]T_f=659.78-290.5[/tex]
[tex]T_f=369.28\ K[/tex]
You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.
(a) What are the two possibilities for its x-component?
(b) Assuming the x-component is known to be positive, specify the magnitude of the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the negative x-direction.
(c) Specify the direction of the vector.
Answer:
Explanation:
a)Magnitude = [tex]\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }[/tex]
84=[tex]\sqrt{(0- (-67))^{2} + (x-0)^{2} }[/tex]
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = [tex]\sqrt{(0- (67))^{2} + (-130.67)^{2} }[/tex] = 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - [tex]Tan^{-1}(67/130.67)degrees[/tex] i.e 152.85 degrees from the +ve x-axis.
A) The two possibilities for the x-component are; +50.67 units or -50.67 units
B) The magnitude of the vector added to the original one is; 146.85 units
C) The direction of the vector is; θ = 207.15°
A) We are given;
Magnitude of vector = 84 units
Y-component of the vector = -67 units
We know that the formula for for 2 vectors like this in x and y direction is;
A = xi^ + yj^
Where A is the magnitude of the resultant
x is the value of the x-component
y is the value of the y-component
Thus;
A = √(x² + y²)
84 = √(x² + (-67)²)
84² = x² + 4489
7056 = x² + 4489
x = ±√(7056 - 4489)
x = ±50.67 units
B) From A above, let us take the positive value of the x-component and as such our original vector will be;
A = 50.67i^ - 67j^
We want to add another vector to this that would make the resultant to be -80 units in the x direction. Thus, A = -80i and if the new additional vector is V^, then we have;
-80i^ = (50.67i^ - 67j^) + V^
V^ = -(80 + 50.67)i^ + 67j^
V = -130.67i^ + 67j^
The magnitude of vector V is;
V = √(x² + y²)
V = √(-130.67)² + 67²)
V = 146.85 units
C) The direction of the vector V is;
θ = tan^(-1) (y/x)
θ = tan^(-1) (67/-130.67)
θ = -27.15°
Since it points entirely in the negative x axis, then the angle is;
180 - (-27.15) = 207.15°
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the volume of an iron sphere is 3.00cm cubed after being heated from 20.0c to 600.0 c what was the initial volume of the iron sphere at 20.0c
Answer:
V = 2.94cm³
The initial volume at 20°C is 2.94cm³
Explanation:
As the temperature of the iron sphere increases the volume of the sphere also increase.
Using the equation for volumetric expansion:
∆V = VαΔT
where ;
V is the initial volume
α is the volumetric expansion coefficient
∆V is the change in Volume
∆T is the change in temperature
After Expansion the final volume can be written as:
Vf = V + ∆V
Vf = V + VαΔT
Vf = V(1 + αΔT)
making V the subject of formula;
V = Vf/(1+αΔT) .....1
Given:
Vf = 3.00cm³
ΔT= 600-20 = 580
And from the test book.
α = 35.5 × 10⁻⁶K⁻¹
Substituting the values into eqn 1
V = 3.00/(1+580× 35.5×10^-6)
V = 3.00/(1+0.021) = 3.00/1.021
V = 2.94cm³
The initial volume at 20°C is 2.94cm³
Answer: 1cm3
Explanation:
V1 =?
V2 = 3cm3
T1 = 20°C = 20 + 273 = 293K
T2 = 600°C = 600 + 273 = 873K
V1 /T1 = V2 /T2
V1 / 293 = 3 / 873
V1 = 293 x ( 3 / 873)
V1 = 1 cm3
The removal, installation, and repair of landing gear tires by the holder of a private pilot certificate on an aircraft owned or operated is considered to be
Answer:
Preventive Maintenance.
Explanation:
Preventive maintenance is nothing but The removal, installation, and repair of landing gear tires by the holder of a private pilot certificate on an aircraft owned or operated. Preventive maintenance in an aircraft is performed by any person with holding at least a private pilot certificate. Only he/she can approve an aircraft to return to service after performing Preventive maintenance tests.
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble simultaneously at opposite ends of a diameter, and stick. W is the turntable's angular velocity, in rpm, just after this event
There are mistakes in the question.The correct question is here
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?
Answer:
w=50 rpm
Explanation:
Given data
The mass turntable M=2kg
Diameter of the turntable d=20 cm=0.2 m
Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s
Two blocks Mass m=500 g=0.5 kg
To find
Turntable angular velocity
Solution
We can find the angular velocity of the turntable as follow
Lets consider turntable to be disk shape and the blocks to be small as compared to turntable
[tex]I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}[/tex]
where I is moment of inertia
[tex]w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm[/tex]
The problem is a case of angular momentum conservation within the domain of rotational dynamics in physics. The turntable's initial angular momentum remains conserved despite the addition of the blocks. By accounting for the added moment of inertia from the blocks, the final angular velocity of the system can be calculated.
Explanation:The subject we're discussing here comes under the physics concept of rotational dynamics particularly focusing on the conservation of angular momentum.
Before the blocks hit the turntable, we know that the turntable is rotating with an angular velocity given in RPM (revolutions per minute), which we can convert to rad/s for our calculations. So, the initial angular momentum can be represented as Lim = (moment of inertia of the system) * (initial angular velocity).
Once the blocks fall onto the turntable, they contribute to the moment of inertia of the system, while the angular momentum of the system remains conserved. Thus resulting in a decreased angular velocity. The final angular momentum can be represented as Lfm = (moment of inertia including the blocks) * (final angular velocity).
Since the initial and final angular momenta need to be equal (Lfm = Lim), we can solve the resulting equation for the final angular velocity.
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A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.
Final answer:
The force on the skier parallel to the slope is found by calculating the component of the skier's weight that acts along the slope. This force is the weight multiplied by the sine of the angle of the slope, which results in 534.8 Newtons for a 103 kg skier on a 32° slope.
Explanation:
To determine the force on the skier parallel to the slope, we can make use of the component of the gravitational force along the slope. Since we are neglecting friction, the only force acting on the skier in the direction parallel to the slope is the component of the skier's weight in that direction.
The weight of the skier can be calculated by multiplying the mass (m) by the acceleration due to gravity (g), which is W = m × g. The component of the weight parallel to the slope is Wparallel = W × sin(θ), where θ is the angle of the slope. Substituting the given values, we have W = 103 kg × 9.8 m/s² = 1009.4 N. The parallel component is then 1009.4 N × sin(32°).
To find the sine of 32°, we use a calculator and get sin(32°) ≈ 0.5299. Multiplying this by the weight gives the parallel force on the skier, which is 1009.4 N × 0.5299 ≈ 534.8 N. Therefore, the force on the skier parallel to the slope is 534.8 Newtons.
A simple pendulum is swinging back and forth through a small angle, its motion repeating every 1.06 s. How much longer should the pendulum be made in order to increase its period by 0.32 s?
Answer:
The pendulum should be made longer by 0.194m in order to increase its period by 0.32s
Explanation:
using the formula T= 2π[tex]\sqrt{\frac{L}{g} }[/tex]
rearranging the equation and making L subject of formula we have;
L=T²g/4π²
lets calculate the length when T=1.06s
g=9.8m/s² , π=3.124
[tex]L=\frac{1.06^{2}*9.8 }{4*3.142^{2} }[/tex]
L=0.279m
the new period after its increased by 0.32s = 1.06+0.32 =1.38s
[tex]L_{2}=\frac{1.38^{2}*9.8 }{4 *3.142^{2} }[/tex]
[tex]L_{2}=0.473m[/tex]
increase in length = 0.473 -0.279
=0.194m
The pendulum should be "0.194 m" longer.
According to the question,
Time,
[tex]T = 1.06 \ s[/tex]We know,
[tex]g = 9.8 \ m/s^2[/tex][tex]\pi = 3.124[/tex]By using the formula,
→ [tex]T = 2 \pi \sqrt{\frac{L}{g} }[/tex]
or,
→ [tex]L = \frac{T^2g}{4 \pi^2}[/tex]
By substituting the values, we get
→ [tex]= \frac{1.06^2\times 9.8}{4\times 3.142^2}[/tex]
→ [tex]= 0.279 \ m[/tex]
Now,
The new period after it increased by 0.32 s, we get
= [tex]1.06+0.32[/tex]
= [tex]1.38 \ s[/tex]
then,
→ [tex]L_2 = \frac{1.38^2\times 9.8}{4\times 3.142^2}[/tex]
[tex]= 0.473 \ m[/tex]
hence,
The increase in length will be:
= [tex]L_2-L[/tex]
= [tex]0.473-0.279[/tex]
= [tex]0.194 \ m[/tex]
Thus the answer above is right.
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A ball has a speed of 15 m/s. Only one external force acts on the ball. After this force acts, the speed of the ball is 7 m/s. Has the force done positive, zero, or negative work on the ball?
Answer:
Negative work
Explanation:
The work-energy theorem states that the work of the resultant forces acting on a particle modifies its kinetic energy:
[tex]W=\Delta K\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\W=\frac{m}{2}(v_f^2-v_0^2)\\W=\frac{m}{2}((7\frac{m}{s})^2-(15\frac{m}{s})^2)\\W=\frac{m}{2}(-178\frac{m^2}{s^2})[/tex]
Since the mass of the ball has to be positive, the work is negative.
The external force did negative work on the ball because the ball's speed decreased from 15 m/s to 7 m/s, indicating a decrease in kinetic energy.
Explanation:To determine whether the force has done positive, negative, or zero work on the ball, we must consider the change in the ball's kinetic energy. Work done by a force is defined as the change in kinetic energy of an object. The formula for work done (W) is given by the change in kinetic energy:
W = ΔKE = ½mv2final - ½mv2initial
Since the speed of the ball decreased from 15 m/s to 7 m/s, the kinetic energy of the ball also decreased. A decrease in kinetic energy means that negative work was done on the ball by the external force.
What conclusion can be derived by comparing the central tendencies of the two data sets?
A: {7, 6, 3, 1, 6, 2, 4, 6, 3, 5}
B: {2, 2, 2, 3, 4, 5, 2, 8, 7, 6}
A.
The mean of set A is smaller than the mean of set B.
B.
The median of set A is greater than the median of set B.
C.
The median and the mean of set B are greater than those of set A.
D.
The mode of set B is greater than the mode of set A.