Answer:
Temperature will be 305 K
Explanation:
We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]
Power absorbed P = 3800 watt
Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]
According to Boltzmann rule power radiated is given by
[tex]P=\sigma AT^4[/tex]
[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]
[tex]T^4=87.0381\times 10^8[/tex]
[tex]T=305K[/tex]
So temperature will be 305 K
Colors seen when gasoline forms a thin film on water are a demonstration of
Final answer:
The colorful effects seen when gasoline creates a thin film on water are a demonstration of thin-film interference, due to light waves reflecting and interfering from the surfaces of the film. The phenomenon causes different colors to appear based on the film's thickness and the light's wavelength.
Explanation:
The colors seen when gasoline forms a thin film on water are due to thin-film interference. This optical phenomenon occurs when light waves reflected from the top and bottom surfaces of a thin film, such as oil or a soap bubble, combine. The interference can be constructive or destructive, depending on the phase difference between the waves. Constructive interference results in bright colors, while destructive interference can cause darkness or cancellation of color. The film's varying thickness results in different colors because each color in the visible spectrum has a different wavelength; thinner or thicker areas interfere with different colors differently.
Early experiments in this area by Agnes Pockels helped to lay the groundwork for our understanding of thin films. Interference effects are most noticeable when light interacts with objects that have sizes similar to the wavelength of light, a principle that is also applied in designing anti-reflection coatings and optical filters.
Final answer:
The colorful effects seen when gasoline or oil creates a thin layer on water are due to thin-film interference, a result of constructive interference of light reflecting off the film's surfaces.
Explanation:
The colors seen when gasoline forms a thin film on water are a demonstration of thin-film interference. This occurs due to the interference of light that is reflected off different surfaces of a thin film such as oil or a soap bubble. The phenomenon is explained by the wave nature of light, which when interacting with objects similar in size to its wavelength, causes some wavelengths of light to interfere constructively, creating bright and vivid colors. The observed colors vary because the thickness of the film affects which wavelengths interfere constructively.
Historical contributions to the study of thin-film interference include the work of Agnes Pockels. She developed a trough for measuring surface films, aiding in the investigation of these colorful effects. Moreover, commercial applications like anti-reflection coatings and optical filters utilize thin film interference principles for effectiveness.
Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 9.00 cm. Find the amplitude of the resultant wave.
Answer:
Amplitude of the resultant wave = 15.72 cm
Explanation:
If two identical waves are traveling in the same direction, with the same frequency, wavelength and amplitude; BUT differ in phase the waves add together.
A = 9cm (amplitude)
φ = 45 (phase angle)
The two waves are y1 and y2
y = y1 + y2
where y1 = 9 sin (kx - ωt)
and y2 = 9 sin (kx - ωt + 45)
y = 9 sin(kx - ωt) + 9 sin(kx - ωt + 45) = 9 sin (a) + 9 sin (b)
where a = (kx - ωt)
abd b = (kx - ωt + 45)
Apply trig identity: sin a + sin b = 2 cos((a-b)/2) sin((a+b)/2)
A sin ( a ) + A sin ( b ) = 2A cos((a-b)/2) sin((a+b)/2)
We have that
9 sin ( a ) + 9 sin ( b ) = 2(9) cos((a-b)/2) sin((a+b)/2)
= 2(9) cos[(kx - wt -(kx - wt + 45))/2] sin[(kx - wt +(kx -wt +45)/2]
y = 2(9) cos (φ /2) sin (kx - ωt + 45/2)
The resultant sinusoidal wave has the same frequency and wavelength as the original waves, but the amplitude has changed:
Amplitude equals 2(9) cos (45/2) = 18 cos (22.5°) = 18 * -0.87330464009
= -15.7194835217 cm ≅ 15.72 cm
since amplitudes cannot be negative our answer is 15.72 cm
The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.
Answer:
Force constant will be 1195.85 N/m
Work done will be 1.6859 J
Explanation:
We have given the force, F = 63.5 N
Spring is stretched by 5.31 cm
So x = 0.0531 m
Force is given , F = 63.5 N
We know that force is given by [tex]F=kx[/tex]
So [tex]63.5=k\times 0.0531[/tex]
k = 1195.85 N/m
Now we have to find the work done
We know that work done is given by
[tex]W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J[/tex]
You are traveling on a narrow two-lane road and are approaching a child riding his bike on the right side of the road. A line of oncoming vehicles following a slow moving truck is approaching in the oncoming lane. You should flash your lights and tap your horn and:___________.
You are travelling on a narrow two-lane road and are approaching a child riding his bike on the right side of the road. A line of oncoming vehicles following a slow moving truck is approaching in the oncoming lane. You should flash your lights and tap your horn and: slow down and move closer to the truck
Explanation:
Change just a single lane at once. While switching to another lane to get ready for a turn, you should flag your goal to do as such at any rate 200 feet preceding evolving lanes or turning. Your sign separation must be in any event 300 feet before the turn in the event that you are working a vehicle in a speed zone of at any rate 50 miles for every hour.
Try not to zigzag all around lanes, which will significantly build your hazard of a mishap. On the thru-way, more slow vehicles should utilise the correct lane. Leave the left-hand lane for quicker moving or passing vehicles. Keep below principles when you are moving to another lane:
Make sure that there is no traffic in front of you in the lane you might want to enter Check your mirrors for any vehicles that are getting ready to pass you Briefly turn your head toward the lane that you are entering to ensure that there is no vehicle in your vulnerable side and that there is adequate space to move into the adjoining lane Use your blinker to caution different drivers of your goal to change lanesSmoothly move into the new driving laneYou should slow down and prepare to stop if necessary, yielding to the child on the bike. Flashing lights and honking the horn are not substitutes for cautious driving. Obey local traffic laws and be patient until it is safe to pass.
You are traveling on a narrow two-lane road and are approaching a child riding his bike on the right side of the road. A line of oncoming vehicles following a slow moving truck is approaching in the oncoming lane. To ensure safety, it's crucial not to rely solely on flashing your lights or tapping your horn. Your primary focus should be on slowing down and preparing to stop if necessary to yield to the child on the bike. Remember, flashing lights and using your horn can alert others, but they are not substitute for cautious driving. It's vital to exercise patience and wait until the oncoming traffic has passed and it is safe to overtake the child with a safe distance. Additionally, as a part of traffic safety, always be aware of and obey local laws regarding passing school buses with flashing lights, driving in school zones, and the general requirement to drive on the right hand side when encountering oncoming traffic.
Suppose that two objects attract each other with a gravitational force of 16 units. if the mass of object one was doubled, and if the distance between the objects was tripled, then what would be the new force of attraction?
Answer: 3.5units
Explanation:
Gravitational force existing between the two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between the masses.
Mathematically, F = GMm/r^2
G is the gravitational constant
M and m are the masses
r is the distance between the masses.
If the force of attraction between the masses is 16units, it becomes,
16 = GMm/r^2... (1)
If the mass of object 1 is doubled and distance tripled, we will have
F= G(2M)m/(3r)^2
F=2GMm/9r^2... (2)
Solving eqn 1 and 2 to get the new Force
Dividing eqn 1 by 2, we have
16/F = GMm/r^2 ÷ 2GMm/9r^2
16/F = GMm×9r^2/r^2×2GMm
16/F = 9/2(upon cancelation)
Cross multiplying we have
9F=32
F= 32/9
F= 3.5units
Answer:
Answer: F = 4 units
Explanation:
If the distance is increased by a factor of 2, then force will be decreased by a factor of 4 (22). The new force is then 1/4 of the original 16 units.
F = (16 units ) / 4 = 4 units
One of the predicted problems due to global warming is that ice in the polar ice caps will melt and raise sea level everywhere in the world. Where is this problem the most worrisome?
Answer: The global warming is affecting even the most cold places on earth, where we can find the polar ice caps and the glaciers. These places belong to the arctic glacial ocean.
Explanation: In the pass of time, due the human action on the enviroment, the desappear of the polar ice cover are increasing day after day. We, people, are loosing the prime earth air condicioner sistem, which helps to regulate the climate sistem since the entire history of humanity.
In the year of 2018, researches estimated that the total volume of Arctic sea ice in late summer declined, by two estimates, by 75 percent in half a century.
We must know that the decrease of the arctic sea ice have deep effects on the climactic sistem of earth and, considering the polar ice caps desappearence, the sea temperature will increase a lot, due the reflection of ultraviolet rays on the blue ocean, (not anymore on the white ice), transforming what should be the earth air condicioner sistem to a strong heater.
Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was approaching the light, it appeared yellow to me because of the Doppler effect. The red light from the traffic signal was shifted up in frequency because I was traveling towards it, just like the pitch of an approaching car rises as it approaches you." a. Calculate how fast Slick Willy must have been driving, in meters per second, to observe the red light (wavelength of 687 nm) as yellow (wavelength of 570 nm). (Treat the traffic light as stationary, and assume the Doppler shift formula for sound works for light as well.)
Answer:
61578948 m/s
Explanation:
λ[tex]_{actual}[/tex] = λ[tex]_{observed}[/tex] [tex]\frac{c+v_{o}}{c}[/tex]
687 = 570 [tex](\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )[/tex]
[tex]v_{o}[/tex] = 61578948 m/s
So Slick Willy was travelling at a speed of 61578948 m/s to observe this.
Answer:
[tex]6.11*10^7m/s[/tex]
Explanation:
The Doppler effect formula for an observer approaching a source is given by equation (1);
[tex]f_o=\frac{f(v+v_o)}{v-v_s}...................(1)[/tex]
where [tex]f_o[/tex] is the frequency perceived by the observer, v is the actual velocity of the wave in air, [tex]v_o[/tex] is the velocity of the observer, [tex]v_s[/tex] is the velocity of the source and [tex]f[/tex] is the actual frequency of the wave.
The actual velocity v of light in air is [tex]3*10^8m/s[/tex]. The relationship between velocity, frequency and wavelength [tex]\lambda[/tex] is given by equation (2);
[tex]v=\lambda f...........(2)[/tex]
therefore;
[tex]f=\frac{v}{\lambda}...............(3)[/tex]
We therefore use equation (3) to find the actual frequency of light emitted and the frequency perceived by Slick Willy.
Actual wavelength [tex]\lambda[/tex] of light emitted is 678nm, hence actual frequency is
given by;
[tex]f=\frac{3*10^8}{687*10^{-9}}\\f=4.37*10^{14}Hz[/tex]
Also, the frequency perceived by Slick Willy is given thus;
[tex]f_o=\frac{3*10^8}{570*10^{-9}}\\f=5,26*10^{14}Hz[/tex]
The velocity [tex]v_s[/tex] of the source light is zero since the traffic light was stationary. Substituting all parameters into equation (1), we obtain the following;
[tex]5.26*10^{14}=\frac{4.37*10^{14}(3*10^{8}+v_o)}{3*10^8-0}[/tex]
We then simplify further to get [tex]v_o[/tex]
[tex]10^{14}[/tex] cancels out from both sides, so we obtain the following;
[tex]5.26*3*10^8=4.37(3*10^8+v_o)[/tex]
[tex]15.78*10^8=13.11*10^8+4.37v_o\\4.37v_o=15.78*10^8-13.11*10^8\\4.37v_o=2.67*10^8[/tex]
Hence;
[tex]v_o=\frac{2.67*10^8}{4.37}\\v_o=6.11*10^7m/s[/tex]
The atoms in a sample of carbon must contain nuclei with the same number of
Answer:
Protons
Explanation:
All the atoms in the carbon sample will contain the same atomic number, represented by Z.
Also, the nucleus of an atom consists of protons and neutron and for the atomic number to be same, the number of protons in the atom must be same.
Therefore, the nucleus can have variation in the number of neutron but not proton for the same carbon sample giving rise to the isotopes of carbon.
Isotopes:
The elements that contains the same atomic number but their mass number is different.
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the minimum μs between the disk and the floor to allow rolling without slipping?
Answer:
[tex]\mu_s=1.0205[/tex]
Explanation:
Given:
mass of solid disk, [tex]m=2\ kg[/tex]radius of disk, [tex]r=2\ m[/tex]force of push applied to disk, [tex]F=20\ N[/tex]distance of application of force from the center, [tex]s=0.3\ m[/tex]For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.
[tex]\therefore F<f_s[/tex]
where:
[tex]f_s[/tex] = static frictional force
[tex]\Rightarrow 20<\mu_s\times N[/tex]
[tex]\Rightarrow 20<\mu_s\times m.g[/tex]
[tex]\Rightarrow 20<\mu_s\times 2\times 9.8[/tex]
[tex]\mu_s>1.0204[/tex]
A weight watcher who normally weighs 400 N stands on top of a very tall ladder so she is one Earth radius above Earth's surface (i.e. twice her normal distance from Earth's center). How much is her weight there? Group of answer choices
Answer:
100 N
Explanation:
Let us first find the mass of the person when they are on the earth's surface
[tex]F=G\frac{Mm}{r^{2} }\\[/tex]
Now let us see how F changes if the distance between two masses is doubled
[tex]F_{1} = G\frac{Mm}{(2r)^{2} } = \frac{1}{4} *G\frac{Mm}{r^{2} }[/tex]
So when the distance of separation is doubled, the force of gravity between the two masses decreases by a factor of four
Therefore her new weight = 400 N/4 = 100 N
Final answer:
The weight of the weight watcher on top of the tall ladder would be 1/4 of her normal weight due to the decrease in gravity with distance from Earth's center.
Explanation:
The weight of the weight watcher on top of a very tall ladder, twice her normal distance from Earth's center, would be 1/4 of her normal weight. This is because the gravitational force decreases with distance from the center of the Earth. When experiencing 1/4 the normal gravitational force, her weight would be 1/4 of the original weight.
What would a plot of p versus 1/v look like for a fixed quantity of gas at a fixed temperature?
Answer:
Straight Line
Explanation:
For an ideal gas,
PV = nRT
For a fixed quantity ( constant number of moles) of a gas at fixed temperature
Right side of the equation will be constant
Thus,
PV = C
So, P = [tex]\frac{C}{V}[/tex] .
Thus P is directly related to [tex]\frac{1}{V}[/tex]
That's why plot between P and [tex]\frac{1}{V}[/tex] will be an straight line.
At what fraction of its current radius would the free-fall acceleration at the surface be three times its present value?
Answer:
0.577
Explanation:
The initial weight is
mg= [tex]\frac{GMm}{r_1^2}[/tex].......................1
and the final weight is
[tex]3mg= \frac{GMm}{r_2^2}[/tex]......................2
now to calculate [tex]r_2/r_1[/tex]
now dividing equation 1 by equation 2 we get
[tex]\frac{1}{r_1^2}/\frac{1}{r_2^2}[/tex]= 1/3
[tex]\frac{r_2}{r_1} =\sqrt{\frac{1}{3} }[/tex]
[tex]\frac{r_2}{r_1}[/tex]=0.577
The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as its length. If a beam one third 1/3 foot wide, one half 1/2 foot high, and 15 feet long can support 30 tons, find how much a similar beam can support if the beam is one half 1/2 foot wide, one third 1/3 foot high, and 15 feet long?
Answer:13.33 tons
Explanation:
Given
Weight of a material varies as
[tex]W\propto b[/tex] (width)
[tex]W\propto h^2[/tex] (height)
[tex]W\propto \frac{1}{L}[/tex] (length)
[tex]W=k\frac{bh^2}{L}[/tex]
Dimension of first beam
[tex]b=\frac{1}{3}[/tex] foot
[tex]h=\frac{1}{2}[/tex] foot
[tex]L=15[/tex] foot
Weight supported [tex]W=20 tons[/tex]
Second beam
[tex]b=\frac{1}{2} foot[/tex]
[tex]h=\frac{1}{3} foot[/tex]
[tex]h=15 feet[/tex]
let weight of second beam be [tex]W_2[/tex]
taking both beams at the same time
[tex]\frac{20}{W_2}=\frac{\frac{1}{3}\times (\frac{1}{2})^2}{15}\times \frac{15}{\frac{1}{2}\times (\frac{1}{3})^2}[/tex]
[tex]\frac{20}{W_2}=\frac{3}{2}[/tex]
[tex]W_2=\frac{40}{3} \approx 13.33 tons[/tex]
A stamping machine begins turning out components that are out of tolerance. The manager removes the machine from service to ensure that more defective components are not produced and notifies maintenance to repair the machine. This is an example of ________. A. immediate corrective action B. benchmarking C. corporate governance D. disciplinary action E. basic corrective action
Answer:
Option A
Explanation:
Immediate corrective measure happens at the instant as it triggers a reaction according to a particular situation.
The time period of the solution is also on the same pattern and is not sustainable.
Thus the case where the stamping machine malfunctions and there is immediate removal of the machine from the service in order to reduce the production of the defective components is an example of immediate corrective measure.
A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom when the gas temperature is 315 ∘C. The air above the piston is at 1.00 atm pressure.
A) What is the gas pressure inside the cylinder?B) What will the height of the piston be if the temperature is lowered to18 ∘C?
Considering the gasses law and all the parameters, 1) the pressure inside the cylinder is 3249.4 Pa, 2) the height of the piston at 18ºC is 0.4452 m.
What is the Law of gasses?
The law of gasses is a group of chemical and physical laws that allows understanding the behavior of gasses in a close system.
The law of gasses considers different parameters, such as
Pressure, P: the amount of force applied on a surface. It is expressed in Pascals (Pa) or atmospheres (atm). 1 atm = 101325 Pa.
Volume, V: spaces occupied by a certain amount of mass. It is expressed in litters (L).
Temperature, T: it is a measure of the particles' internal agitation. It is expressed in kelvins (K). One centigrade = 1 + 273 kelvins.
When talking about standard conditions in gas, we refer to 1 atm pressure, 273 K of temperature (0ºC), and 22.4 L/mol of volume.
The different laws are,
Boyle law ⇒ P₁V₁ = P₂V₂ ⇒ P and V are inversely proportional. T is constant.Charles law ⇒ V₁/T₁ = V₂/T₂ ⇒ V and T are directily porportional. P is constant.Gay-Lussac law ⇒ P₁/T₁ = P₂/T₂ ⇒ T and P are directily porportional. V is constant.The relationship between these three laws and the variables is as follows,
(P₁V₁) / T₁ = (P₂V₂) / T₂
According to this framework, we can answer the questions in this porblem.
Available data:
D = 24-cm² = 0.0024 m²m = 15 kgh = 90 cm = 0.9 mT₀ = 315ºC ⇒ 315 + 273 = 588 K.Pair = 1atmA) What is the gas pressure inside the cylinder?
We know that P is defined by the relationship between force and area.
P = F/A
So, let us define Force and Area.
FORCE
F is the force applied by the piston. So we need to get the F of the piston.
To do it, we will use the following formula,
W = mg ⇒ Where m is mass (15 kg) and g is gravity (9.8m/s²)
Since W is a type of force, we can replace this value in the previous general formula,
P = F/A = W/A = mg/A
AREA
Area A is defined by the following formula
A = π r²
Where
- π = 3.1416
- r = D/2
Now, we can replace A in the general formula as follows,
P = mg/A = mg / π (D/2)²
The pressure inside the cylinder is,
P = mg / π (D/2)²
P = (15 x 9.8) / 3.1416 (12)²
P = 147 / 452.39
P = 0.32494
P = 3249.4 Pa
The gas pressure inside the cylinder 3249.4 Pa.
B) What will the height of the piston be if the temperature is lowered to 18 ∘C?
T₁ = 588 K
T₂ = 18 ºC ⇒ 18 + 273 = 291 K
Now, volume can be calculated as follows,
V = π r² h = A h
Where
- π = 3.1416
- r = D/2 = 12
- h₁ = 0.9 m
- A = 452.39 cm² = 0.4524 m²
Replacing,
V₁ = A h₁
V₁ = 0.4524 m² x 0.9 m
V₁ = 0.4071 m³
Now, we have the initial volume V₁, and the inicial and final temperatures, T₁ and T₂. We will use this values and the Charles law to get the final volume, V₂.
V₁/T₁ = V₂/T₂
V₂ = T₂ (V₁/T₁)
V₂ = 291 (0.4071/588)
V₂ = 291 (0.4071/588)
V₂ =0.2014
Now that we have the final volume, we can use this value to get the final height.
V₂ = A h₂
h = V₂/A
h₂ = 0.2014 m³ / 0.4524 m²
h₂ = 0.4452 m
The height of the piston when the temperature is lowered to 18ºC is 0.4452 m.
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Final answer:
Starting with Part A, the gas pressure inside the cylinder is calculated using the piston's weight and area of the cylinder's cross-section, plus atmospheric pressure. For Part B, the height of the piston at 18 °C is found using the combined gas law, with the final volume corresponding to the new height under constant pressure.
Explanation:
To answer the student's questions regarding the cylinder and piston system, we must use the principles of gas laws and mechanics.
Part A: Gas Pressure Inside the Cylinder
The pressure exerted by the piston onto the gas can be calculated using the formula P = F/A, where F is the force and A is the area. The force due to the piston's weight (F) is the mass of the piston times the gravitational acceleration (F = m * g). The area (A) is the cross-sectional area of the cylinder, which is πr² for a circle. Adding the atmospheric pressure above the piston gives us the total pressure exerted on the gas inside the cylinder.
Part B: Height of the Piston at 18 °C
Assuming the process is isobaric (constant pressure), the change in height can be found using the combined gas law which relates pressure, volume, and temperature of a gas. In this case, P1V1/T1 = P2V2/T2, where V1 and V2 are the initial and final volumes, respectively, determined by the product of the cross-sectional area and the height of the piston, P is pressure, and T is temperature in Kelvin. We are looking for the final volume, and thus the final height of the piston when the temperature is lowered.
A lift (elevator) is operated by an electric motor. It descends from the 10th floor to the 2nd floor at a constant speed. What is the main energy transformation during this journey?
#1
gravitational potential energy → kinetic energy.
#2
electrical energy → kinetic energy.
#3
kinetic energy → thermal energy.
#4
electrical energy → thermal energy.
Answer:
#4 Electrical energy to thermal energy
Explanation:
You clue there is constant speed, this means that the lift is not accelerating. So if that is the case, the lift is experiencing a resistive force against gravity. The one doing exerting the resistive force is the motor, which runs on electrical energy. Resistive force transforms into heat, so that makes it thermal energy. And so, the answer would be the last case.
Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one place and climb to the top of 2 m high ladders. How muchhas the center of mass of the Earth (mE = 5.90×1024 kg) displaced as a result?
The center of mass of the Earth would be displaced by approximately [tex]\(7.65 \times 10^{-15}\)[/tex] meters.
Explanation.When approximately 1 billion people in China, each with an average mass of 55 kg, climb 2 meters high on ladders, the displacement of the Earth's center of mass can be calculated using the concept of center of mass. Considering the mass of the people and the distance they moved, the displacement is a minuscule [tex]\(7.65 \times 10^{-15}\)[/tex]meters. This displacement is incredibly small relative to the Earth's overall size and mass, indicating that while the collective movement of a large number of people does have an effect, it's an exceedingly tiny one.
Even with such a substantial number of individuals moving, the impact on the Earth's center of mass is practically negligible due to the massive scale and weight of the planet. The displacement is a fascinating mathematical concept illustrating the relationship between mass distribution and movement, showcasing how even large numbers of individuals can exert an effect that remains imperceptible on the scale of the entire Earth.
This answer highlights how a considerable number of individuals moving can have a collective effect on the Earth's center of mass, but due to the immense size and mass of the planet, this displacement is minuscule and practically undetectable.
Jet streams can form near Earth’s surface as well. One such jet develops just above the central plains of the United States, where it occasionally attains speeds exceeding 60 knots. This wind speed maximum, which usually flows from the south or southwest, is known as a(n) ____________________ jet.
Answer:
Low level jet
Explanation:
Jet streams can also form close to the earth's surface. of all such type of jet one jet grows just over the main central plains of United States, where speeds approaching 60 knots are rarely reached. This high point wind speed jet is recognised as low level jet, which typically comes from south or southwest.
As title indicates, , it termed as a rapidly-moving stream of air in low atmosphere levels .
A skier of mass 71 kg is pulled up a slope by amotor-driven cable.The acceleration of gravity is 9.8 m/s2.How much work is required to pull him 62 mup a 31◦slope (assumed to be frictionless) ata constant speed of 2.58 m/s?Answer in units of kJ
Answer:
Work done will be 22216.894 J
Explanation:
We have given mass of the skier m = 71 kg
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Slop is given as [tex]\Theta =31^{\circ}[/tex]
Distance d = 62 m
So vertical distance [tex]d_y=dsin\Theta =62\times sin31^{\circ}=31.932m[/tex]
We know that work done is given by
[tex]W=mgh=mgd_y=71\times 9.8\times 31.93=22216.894J[/tex]
Water is flowing through a pipe whose cross sectional area at point A is larger than the cross sectional area at point B. The reading on a pressure gauge connected to the pipe at point A will be lower than the reading on a pressure gauge connected to the pipe at point B.
a.True
b.False
Answer:
a. True
Explanation:
Pressure is always inversely proportional to the cross section area
This can be proven from the equation P = F/A
Where P = Pressure, F = Force and A = Cross-sectional area
This can also be similarly proven from the continuity equation given as
A₁V₁ = A₂V₂
Where A₁ = Area in first part of pipe (point A)
A₂ = Area in second part of pipe (point B)
V₁ = Velocity of water in first part of pipe (point A)
V₂ = Velocity of water in second part of pipe (point B)
From the continuity equation we can have that
V₁/V₂ = A₂/A₁
Thus the velocity of the liquid at any point in the pipe is inversely proportional to the cross-sectional area of the pipe. The liquid will be moving slowly where the area is large and will be moving rapidly where the area is small, and since the velocity is directly proportional to the force (F = m × v/t)which id in turn directly proportional to the pressure (P = F/A), this proves my answer to be True!
A pair of alkaline D batteries can provide about 98000 J of electric energy at 3.0 V. If a current of 2 A flows through two D batteries while they’re in the circuit of a flashlight, how long will the batteries be able to provide power to the flashlight?
Answer:
4.53 hours
Explanation:
Energy, E = 98000 J
Voltage, V = 3 V
Current, i = 2 A
Let t be the time for which the batteries give the power.
Energy = V x i x t
98000 = 3 x 2 x t
t = 16333.33 seconds
t = 4.53 hours
HELP MEEEEEEEEEEEEEE!!!!!!!!!!!!!!!!!! Please show work! Margaret, whose mass is 52 kg, experienced a net force of 1750 N at the bottom of a roller coaster loop during her school's physics field trip to Six Flags. What is her acceleration at the bottom of the loop?
Answer:
43.46 m/[tex]s^{2}[/tex]
Explanation:
Two forces will be acting on her at the bottom point of the rollar coaster,
One will be mg (vertically downwards) and other will be ma(vertrically upwards) due to the centripetal force .
So, taking the resultant of the forces , we will get
ma-mg = 1750 (Given)
m(a-g) = 1750
52(a-g) = 1750
a-9.81 = 33.65
a = 43.46
In the two-slit experiment, for the condition of bright fringes, the value of m = +2 corresponds to a path difference of λ.
True or False?
Answer:
False
Explanation:
We know that if path difference is even multiple of wavelength then bright fringes are formed and if path difference is odd multiple of wavelength then dark fringes are formed .
For bright fringes
Path difference Δx = m λ
m = 0 , 2 , 4 , 6.......
If m = 2 then the path difference will be
Δx = 2 λ
therefore the above statement if false.
False
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car is travelling at 14.5m/s and only the friction from the tires on the roadway stop the vehicle. Find the coefficient of friction
Answer:
Coefficient of friction will be 0.587
Explanation:
We have given mass of the car m = 500 kg
Distance s = 18.25 m
Initial velocity of the car u = 14.5 m/sec
As the car finally stops so final velocity v = 0 m/sec
From second equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]0^2=14.5^2+2\times a\times 18.25[/tex]
[tex]a=-5.76m/sec^2[/tex]
We know that acceleration is given by
[tex]a=\mu g[/tex]
[tex]5.76=\mu\times 9.81[/tex]
[tex]\mu =0.587[/tex]
So coefficient of friction will be 0.587
Two technicians are discussing the need for the history of the vehicle. Technician A says that an accident could cause faults due to hidden damage. Technician B says that some faults could be related to a previous repair. Which technician is correct?a. Tech A onlyb. Tech B onlyc. Both a and bd. Neither a and b
Answer:
(C) both tech A and tech B
Explanation:
both technicians are correct because an accident could cause faults due to hidden damages which might not have been detected at the time of the accident and some faults could be related to previous repair probable due to human error from the repair man or even improper repair due to lack of adequate knowledge on the vehicle.
A 350-g air track cart on a horizontal air track is attached to a string that goes over a frictionless pulley with a moment of inertia of 6.00 × 10-6 kg ∙ m2 and a radius of 1.35 cm. If the string is pulled vertically downward by a force of 2.50 N,
(a) what is the magnitude of the acceleration of the cart?
(b) what is the tension in the horizontal string between the pulley and the cart?
The acceleration of the cart is 7.14 m/s^2. The tension in the string between the pulley and the cart is approximately 2.26 N.
Explanation:This physics problem involves the principles of dynamics and rotational motion.
(a) First, using Newton's second law, the acceleration a of the cart (mass, m) pulled by a force F is given by a = F/m. So, the acceleration is a = 2.50 N / 0.350 kg = 7.14 m/s^2.
(b) Considering the pulley, the torque τ exerted by the tension in the string is τ = r * T, r being the radius of the pulley. Given that τ = I * α and α = a/r (α being the angular acceleration), we can express the tension T as T = (I * a)/(r^2) = (6.00 × 10^-6 kg*m^2 * 7.14 m/s^2) / (0.0135 m)^2 = approximately 2.26 N.
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A playground ride consists of a disk of mass M = 49 kg and radius R = 1.7 m mounted on a low-friction axle. A child of mass m = 29 kg runs at speed v = 2.6 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.
Answer:
The angular speed is 0.83 rad/s.
Explanation:
Given that,
Mass of disk M=49 kg
Radius = 1.7 m
Mass of child m= 29 kg
Speed = 2.6 m/s
Suppose if the disk was initially at rest , now how fast is it rotating
We need to calculate the angular speed
Using conservation of momentum
[tex]m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}[/tex]
[tex]mvR=(mr^2+\dfrac{Mr^2}{2})\omega[/tex]
Put the value into the formula
[tex]29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}[/tex]
[tex]\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}[/tex]
[tex]\omega_{f}=0.83\ rad/s[/tex]
Hence, The angular speed is 0.83 rad/s.
The angular speed is mathematically given as
wf=0.83rad/s
What is the angular speed?Question Parameter(s):
Generally, the equation for the conservation of momentum is mathematically given as
[tex]m\omega_{i}=(mr^2+\frac{Mr^2}{2})\omega_{f}[/tex]
Therefore
[tex]29\*2.6*1.7=(29\times1.7^2+\frac{49*1.7^2}{2})\omega_{f}[/tex]
wf=0.83rad/s
In conclusion, the angular speed is
wf=0.83rad/s
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A monochromatic light beam is incident on a barium target that has a work function of 2.50 eV. If a potential difference of 1.00 V is required to turn back all the ejected electrons, what is the wavelength of the light beam?a) 355 nmb) 497 nmc) 744 nmd) 1.42 pme) none of those answers
By using the concepts of the Photoelectric effect, the wavelength of the light beam incident on a barium target with a work function of 2.50 eV and requiring a potential difference of 1.00V to turn back all the ejected electrons is calculated to be 355 nm.
Explanation:To answer this question, we need to utilize the Photoelectric effect, which explains the relation between light and electrons. It determines the minimum amount of energy that a particle of light, a photon, must possess to eject an electron from a material, which is called the work function. In this case, the work function (W) of the barium target is given as 2.50 eV.
But we also need to account for the potential difference of 1.00V required to turn back the electrons. When 1 eV is required to turn back the electrons, it implies that the electrons have an extra energy value of 1 eV after overcoming the work function. We can combine these energies to find the total energy of the light beam as W + 1.00 eV which gives us 3.50 eV.
Now, we convert the total energy of the light beam from electronvolts to joules by multiplying with a conversion factor of 1.6 x 10^-19 J/eV. Therefore, we get E = 3.50 eV x 1.6 x 10^-19 J/eV = 5.6 x 10^-19 J.
We can then find the frequency of the light beam using the equation E = hv, where h is Planck’s constant (6.63 x 10^-34 Js) and v is the frequency. So, v = E / h = (5.6 x 10^-19 J) / (6.63 x 10^-34 Js) = 8.44 x 10^14 Hz.
Finally, we calculate the wavelength using the equation c = λv, where c is the speed of light (3 x10^8 m/s), λ is the wavelength, and v is the frequency. So, λ = c / v = (3 x 10^8 m/s) / (8.44 x 10^14 Hz) = 3.55 x 10^-7 m or 355 nm.
Therefore, the wavelength of the light beam is 355 nm.
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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, the ball will break through the strings and keep going. The racket is a potential-energy barrier whose height is the energy of the slowest string-breaking ball. Suppose that a 100 g tennis ball traveling at 200 mph is just sufficient to break the 2.0-mm-thick strings. Estimate the probability that a 120 mph ball will tunnel through the racket without breaking the strings. Give your answer as a power of 10 rather than a power of e.
Answer:
Probability of tunneling is [tex]10^{- 1.17\times 10^{32}}[/tex]
Solution:
As per the question:
Velocity of the tennis ball, v = 120 mph = 54 m/s
Mass of the tennis ball, m = 100 g = 0.1 kg
Thickness of the tennis ball, t = 2.0 mm = [tex]2.0\times 10^{- 3}\ m[/tex]
Max velocity of the tennis ball, [tex]v_{m} = 200\ mph[/tex] = 89 m/s
Now,
The maximum kinetic energy of the tennis ball is given by:
[tex]KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J[/tex]
Kinetic energy of the tennis ball, KE' = [tex]\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s[/tex]
Now, the distance the ball can penetrate to is given by:
[tex]\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}[/tex]
[tex]\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js[/tex]
Thus
[tex]\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}[/tex]
[tex]\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}[/tex]
[tex]\eta = 1.52\times 10^{-35}\ m[/tex]
Now,
We can calculate the tunneling probability as:
[tex]P(t) = e^{\frac{- 2t}{\eta}}[/tex]
[tex]P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}[/tex]
[tex]P(t) = e^{-2.63\times 10^{32}}[/tex]
Taking log on both the sides:
[tex]logP(t) = -2.63\times 10^{32} loge[/tex]
[tex]P(t) = 10^{- 1.17\times 10^{32}}[/tex]
A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 5.00 m from her.
Answer:
1.995 m
Explanation:
Distance of penny as seen by the person = 5 m
Height of person from water surface = 3.50 m
Apparent depth of penny = 5 - 3.50 = 1.5 m
refractive index of water, n = 1.33
real depth / apparent depth = n
real depth = 1.33 x 1.5 = 1.995 m
Thus, the actual depth of water at that point is 1.995 m.