The coefficient of aerial expansion for metal b is one-fourth of the coefficient of aerial expansion for metal a.
Note: Most probably your question was to find the relation between the coefficient of aerial expansion for metal a and metal b.
Coefficient of linear expansion:
The ratio of the change in length of a solid to its original length when there is a unit change in temperature at constant pressure is called the coefficient of linear expansion. It is given by the formula,
α=ΔL/(Lo*ΔT)
where α is the coefficient of linear expansion, ΔL is the change in length Lo is the original length, and ΔT is the change in temperature.
Coefficient of aerial expansion:
The ratio of the change in surface area of a solid to its original surface area when there is a unit change in temperature at constant pressure is called the coefficient of aerial expansion. It is given by the formula,
β=ΔA/(Ao*ΔT)
where β is the coefficient of aerial expansion, ΔA is the change in surface area, Ao is the original surface area, and ΔT is the change in temperature.
Relation between α and β:
The relation between the coefficient of linear expansion and aerial expansion is given by the formula,
β=2α
Given that metal a has a coefficient of linear expansion four times that of metal b. It can be written.
α₁=4α₂
where α₁ is the coefficient of linear expansion for metal a and α₂ is the coefficient of linear expansion for metal b.
Then from the above relation, the ratio of the coefficient of aerial expansion β₁ and β₂ for metal a and metal b respectively is,
β₁/ β₂ = 2*(α₁)÷(2*(α₂))
β₁/ β₂= 2*(4α₂)÷(2*(α₂))
β₁/ β₂=4
β₂= β₁/4
Therefore, the coefficient of areal expansion of metal b is one-fourth of the coefficient of aerial expansion of metal a.
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The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800. Select the correct answer.
The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg
Data obtained from the questionYear 1800 percent = 0.028%Year 2013 percent = 0.040%Mass of air = 500 KgDifference =?How to determine the mass of CO₂ in 500 Kg in year 1800Year 1800 percent = 0.028% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.028% × 500
Mass of CO₂ = 0.14 Kg
How to determine the mass of CO₂ in 500 Kg in year 2013Year 1800 percent = 0.040% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.040% × 500
Mass of CO₂ = 0.2 Kg
How to determine the differenceMass of CO₂ in year 1800 = 0.14 KgMass of CO₂ in year 2013 = 0.2 KgDifference =?Difference = mass in 2013 - mass in 1800
Difference = 0.2 - 0.14
Difference = 0.06 Kg
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Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines
While moving over a separate electric field line, however, has a non-zero value. If the test charge is transferred in a direction perpendicular to E, no work is done and the electric potential does not change. A smooth equipotential surface characterizes every such path.
Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines?Generally, The equation of the potential difference is mathematically given as
[tex]V = Ed cos \theta.[/tex]
Where
[tex]\theta=Angle[/tex]
E=electric field
d= the varing distance
When the charge to be tested is transferred via an electric field line. Since =0, the maximum possible velocity (Vmax) is calculated.
In conclusion, Since no work is done when the test charge is shifted in a direction perpendicular to E, the electric potential does not shift in this case. Any such route is a smooth equipotential surface.
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Which statement is true of a piece of ice at 0°C that is put into a freezer at
-18°C?
The statement which is true about a piece of ice at 0°C which is put into a freezer at -18°C is it having the temperature of the freezer.
What is Temperature?This is referred to the degree of hotness or coldness of a body and the unit is Celsius or Kelvin.
The ice at 0°C will experience a change in temperature of the freezer when put in it in this scenario.
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A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory?
The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s
What is Trajectory ?When any object is thrown from horizontal at an angle θ except 90°, then the path followed by it is called trajectory, the object is called projectile and its motion is called projectile motion.
Here, we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v = 0.
Given :
Horizontal speed= 30m/sVertical Speed= 30 m/sThis problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use
vy = vy0 + ayt
Plug in our givens
0 = 30 - 10t
solve for t
t = 3 seconds
Hence, The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s
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A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a distance of
75.0 m. If the vehicle's acceleration remained constant during the braking, what was the vehicle's initial speed?
Since acceleration is constant, the average and instantaneous accelerations are the same, so that
[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]
By the same token, we have the kinematic relation
[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]
where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.
Substitute everything you know and solve for [tex]v_i[/tex] :
[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]
[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]
[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]
[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]
Dinesh takes 20 minutes to cover a distance of 5 km on a bicycle. Calculate his average speed in.
A) 1m/s
B) 2km/h
Please explain in detail.
Answer:
Soln:Here
Given,
Distance(s)=5m
Time taken(t)=20min=2400seconds
Speed(s)=?
We know that,
Speed(s)=Distance/timetaken
=5m/2400sec
580m/s
Thus the average speed of his bicycle is 580 m/s..
Thank you...
A mass is placed on a smooth inclined plane with an angle of 37e to the horizontal. If theinclined plane is 5.0-m long, how long does it take for the mass to reach the bottom of theinclined plane after it is released from rest?
The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
Height of the inclined plane
The height of the inclined plane is calculated as follows;
h = L(sin 37)
where;
h is height of the planeL is length of the planeh = 5 x sin(37)
h = 3.01 m
Time of motion of the masst = √(2h/g)
t = √(2 x 3.01 / 9.8)
t = 0.78 s
Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
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An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h.Compare it's centripetal acceleration with the acceleration due to gravity.
Answer:
900 km/hr = 9E5 m / 3600 sec = 250 m/sec
ah = v^2 / R = 250^2 m/s / 1000 m = 62.5 m/s^2 horizontal acceleration
av = 9.8 m/s^2 vertical acceleration (due to gravity)
ah / av = 62.5 / 9.8 = 6.38 centripetal acceleration is greater
1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.
Answer:
final velocity=u+at 15+2×5=25
Why should a scientist not be concerned when a fellow scientist responds to their conclusions with skepticism?
The work that is done when twice the load is lifted twice the distance is _______. View Available Hint(s)for Part A The work that is done when twice the load is lifted twice the distance is _______. the same three times as much twice as much four times as much
The work that is done when twice the load is lifted twice the distance is
four times as much
The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.
when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.
Hence The work that is done when twice the load is lifted twice the distance is four times as much
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You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The frequency of the other is:
The frequency of the other wave is 613 Hz or 607 Hz.
The difference between the frequencies of two waves is called the beat frequency.
Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.
Which has a higher frequency is not mentioned. Therefore, there are two possibilities.
Δf = | 610 - 613 | = 3
or
Δf = | 610 - 607 | = 3
Therefore, the frequency of the other wave is 613 Hz or 607 Hz.
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Which class of hazards is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects
Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects.
There are different classes of Hazards
Class 1 - Explosives
Class 2 - Gases
Class 3 - Flammable liquids
Class 4 - Flammable solids
Class 5 - Oxidizers
Class 6 - Toxic materials
Class 7 - Radioactive materials
Class 8 - Corrosive materials
Class 9 - Miscellaneous dangerous goods
Any substance or item, including a gadget, that is intended to function by explosion or that, through a chemical reaction inside itself, is capable of functioning similarly even if not intended to function by explosion, falls within the category of explosive materials (class 1).
Hence, Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards.
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A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their velocities remain constant during the chase?
The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.
What time will the dog catch the rabbit?The time that the dog will catch up with the rabbit is given as follows:
Let the distance covered by the rabbit be x.
Distance covered by dog = x + 30
Time taken = distance/speedThe time taken will be the same T
Time taken by dog, T = (x + 30)/10Time taken by rabbit, T = x/5Equating both times.
(x + 30)/10 = x/5
x = 30 m
Solving for T in equation (ii);
T = 30/5 = 6 minutes
In conclusion, time is obtained as a ratio of distance and speed.
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A brightfield microscope equipped with a ____________ ___________ stop is able to block light from the illuminator creating bright objects on a dark background.
Answer: A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.
Explanation: To find the correct statement about Brightfield microscope, we need to know more about the Brightfield microscope.
what is Brightfield microscope?The Brightfield microscopic technique is generally used with the compound microscopes and, these are one, among the widely used optical illumination techniques. The Brightfield microscope or Compound light microscope is an optical instrument, which uses light rays and produces dark images with a bright background.How Bright field microscope works?A brightfield microscope equipped with a opaque light stop, will produce bright objects on a dark background by blocking the light illuminated from the light source. This opaque light stop, blocks the light rays that arising from the illuminator to the objective lenses and, only allows the reflected or transmitted light of the specimen. As a result, we get bright objects with a dark background.Thus, we can conclude that, A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.
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If f is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?.
The net force acting on test charge is resultant of two force i.e., √2 F
The magnitude of force on test charge due to one charge Q is:
F = k qQ / r²
where k = 9 × 10^9
Similarly the force due to second charge will be:
F = k qQ / r²
Now both the charges of magnitude Q is present and it is pointed at right angle to each other then the force will be resultant i.e.,
F = √ F² + F² = F√2
Hence, net force acting on test charge is √2 F.
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Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the: Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle: Mass of sand Mass of water that filled the space above the sa Volume of the sand: Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the : Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle : Mass of sand Mass of water that filled the space above the sand
Volume of the sand :
The values of the density, volume and masses are:
mass of water that completely filled the bottle = 23.2 gvolume of water that completely filled the bottle = 23.2 cm³volume of the density bottle = 23.2 cm³mass of sand = 24.4 gmass of water that filled the space above the sand = 14.8 gvolume of sand = 8.4 cm³density of sand = 2.9 g/cm³What is density?Density is defined as the ratio of mass and volume of a substance.
Density = Mass/volumeThe given data is as follows:
Mass of empty density bottle = 43.2 g
Mass of density bottle full of water = 66.4 g
Mass of density bottle with some sand = 67.5 g
Mass of density bottle with the sand
filled up with water = 82.3 g:
(a) mass of water that completely filled the bottle = 66.4 - 43.2
mass of water that completely filled the bottle = 23.2 g
(b) volume of water that completely filled the bottle = mass/density
density of water = 1 g/cm³
volume of water that completely filled the bottle = 23.3 g/1 g/cm³
volume of water that completely filled the bottle = 23.2 cm³
(c) volume of the density bottle = 23.2 cm³
(d) mass of sand = 67.5 g - 43.2 g = 24.4 g
(e) mass of water that filled the space above the sand = 82.3 - 67.5 = 14.8 g
(f) volume of sand = volume of bottle - volume of water
volume of water = 14.8 cm³
volume of sand = 23.2 cm³ - 14.8 cm³
volume of sand = 8.4 cm³
(g) density of sand = 24.4 g/8.4 cm³ = 2.9 g/cm³
Therefore, the density of sand is obtained by taking the ratio of the mass of the sand and the volume if the sand.
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PLEASE HELP! An airplane is moving at a velocity of 90.1 m/s and requires 1435 m to reach that velocity. Determine the acceleration of the plane and the time required to reach this velocity.
The time required to reach the velocity is 15.9s.
The acceleration of the plane is 5.67m/s²
Explanation:Acceleration is solved by the given formula.
Acceleration = velocity ÷ Time
Finding Time
Time = distance ÷ speed
Time = 1435m ÷ 90.1m/s = 15.9s
Finding Acceleration
Acceleration = velocity ÷ time
Acceleration = 90.1m/s ÷ 15.9s = 5.67m/s²
A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp.
a) 32%
Ob) 54.65 %
O c) 46.4%
d) 54.6 %
D. The efficiency of the ramp is 54.6 %.
Velocity ratio of the rampV.R = distance moved by effort/distance moved by load = L/h
V.R = (6.12)/(1.53) = 4
Mechanical advantage of the rampM.A = Load/Effort
M.A = (451 x 9.8)/(2025)
M.A = 2.183
Efficiency of the rampE = (M.A / V.R) x 100%
E = (2.183 / 4) x 100%
E = 54.6 %
Thus, the efficiency of the ramp is 54.6 %.
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Which hrase describes what atoms are made of? OA. A cloud of protons and neutrons surrounded by a nucleus of electrons OB. A nucleus of neutrons and electrons surrounded by a cloud of protons OC. A nucleus of protons and neutrons surrounded by a cloud of electrons OD. A nucleus of protons and electrons surrounded by a cloud of neutrons
Answer:
(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons
What is the gravitational potential energy stored in a rock of 100 kg
mass, placed on top of a hill, which is 100 m high?
O 45 kJ
O 98 kJ
O 50 kJ
O 100 kJ
Suppose you put very large-diameter tires on your car. Then, your speedometer will show a speed that is
Answer:
The speed registered by the speedometer depends on the rotations of the axle to which it is connected -
Larger diameter tires will contribute to a larger distance traveled in one rotation of the axle and hence a larger speed and the speed shown on the speedometer will be less than the actual speed.
The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the average depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
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What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm
The magnitude and sign of the charge are 0.8 MC and negative respectively.
To find the answer, we need to know about the electric potential of a point charge.
What's the mathematical expression of the electric potential of a point charge?Mathematically, the electric potential at a distance 'r' from a point charge 'q' is given as (Kq)/r. K is the electrical constant with value 9×10^(-9) in vaccum. What will be the magnitude and sign of a charge, if potential is -3.50V at 2mm?From the expression of electric potential, charge is
q= (potential ×r)/K
= (-3.5×0.002)/ (9×10^(-9))
= -0.8 mega coulomb.
Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.
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A car approaches a stationary police car at 36 m/s. The frequency of the siren (relative to the police car) is 500 Hz. What is the frequency (in Hz) heard by an observer in the moving car as he approaches the police car
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
The Doppler Effect equation says
f = f0 c+v / c
Given
f0 = 500 Hz
c = 343 m/s
v = 36 m/s
by plugging the given data, we get
500 (343+36) / 343 = 552 Hz
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
Doppler effect is the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by the relative motion of the observer and the wave source.
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The eyes of a farsighted person are Blank______ than normal; therefore light rays tend to focus Blank______ the retina.
The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.
What is Farsightedness:
It is also known as hyperopia, is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness is caused by a cornea that isn't curved enough or It causes by an eyeball that's too short.
These two problems prevent light from focusing directly on the retina.
so, the light rays focus behind the retina.
Hence,
The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.
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At which points on the roller coaster is the car not moving?
A
A and E
B
B and C
C
C and D
D
D and E
The maximum potential energy of the car or the when car is not moving at points A and E; option A.
What is potential energy?Potential energy is the energy of a body due to its state or position.
In the rollercoaster motion shown, the maximum potential energy occurs when the car is no longer moving.
At different points other than at maximum potential energy, the energy is a combination of potential and kinetic energy.
The maximum potential energy occurs at A and E, at which point the car is not moving.
In conclusion, at maximum potential energy, the car is not moving.
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A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V. At what voltage will the resistor dissipate 10.0 W
50 V to dissipate a power of 10 W
Given
A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.
Resistor and Voltage The difference in charge between two places is known as voltage. The passage of charge is measured as current. A material's propensity to oppose the flow of charge is known as resistance (current).Ohm's Law describes the relationship between current, voltage, and resistance. This asserts that, given the temperature is constant, the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance.Explanation[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]
10 v provides 2 W of power whereas V provides 10 W of power.
10 V = 2 W
x V=10 W
By using cross multiplication
2 x = 100
x = 50 V
So, 50 V to dissipate a power of 10 W
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Throw two balls from the same height at the same time at an initial speed of 20 m/s. One is thrown vertically down, while the other is thrown vertically up. We want to find the time difference between their landing.
The time difference between their landing is 2.04 seconds.
Time of difference of the two balls
The ball thrown vertical upwards will take double of the time taken by the ball thrown vertically downwards.
Time difference, = 2t - t = t
t = √(2h/g)
where;
h is the height of fallg is acceleration due to gravityApply the principle of conservation of energy;
¹/₂mv² = mgh
h = v²/2g
where;
v is speed of the ballh = (20²)/(2 x 9.8)
h = 20.41 m
Time of motiont = √(2 x 20.41 / 9.8)
t = 2.04 s
Thus, the time difference between their landing is 2.04 seconds.
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Which equation can be used to solve for the magnitude of the velocity (v) at which Frances slides backwards after giving the curling stone a push in Test 3?
A
0 = –(45 kg • v) + (20 kg • 3 m/s)
B
0 = –[(45 kg + 20 kg) • v]
C
60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s)
D
195 kg • m/s2 = (45 kg + 20 kg) • v
A
A 0 = –(45 kg • v) + (20 kg • 3 m/s)
Not moving, means momentum is zero. Thus, momentum after = momentum before (& that momentum is zero)
V = 4/3m/s
(-45 x 4/3) + (20x3) = 0
Solving to find v :
0. = –(45 kg • v) + (20 kg • 3 m/s)
= - (45 kg • v) + 60
-60
-60 = - 45 x v
÷ — 45
4/3 = v
(4/3 = 1.33333333333)
Hope this helps!
The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).
Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.
What is velocity?A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.
A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").
Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.
Therefore, Thus, option C is correct.
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