Tycho Brahe made contributions to astronomy by making observations and gathering data that were later used by Kepler to develop laws of planetary motion.
Another one of the answer options in this group is creating the telescope. Kepler's first law states that all planets have elliptical orbits around the sun with the sun at one of the foci. Perihelion is the name given to the point when a planet is closest to the sun, and aphelion to the point when it is farthest away. A telescope is a tool used to observe far-off objects using their electromagnetic radiation emission, absorption, or reflection. originally exclusively referring to an optical device that observes far-off objects using lenses, curved mirrors, or a combination of the two.
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a fighter jet is traveling at 525 m/s directly away from a communication antenna that broadcasts at 410 mhz . What change in frequency does the fighter jet observe
The fighter jet notices a frequency change of 718 Hz.
from Doppler's effect, f' = f(1 - [tex]\frac{u}{c}[/tex])
Here, c stands for the speed of light, u for the speed of a fighter jet, and f for the antenna's broadcast frequency.
so, f' = f(1 - [tex]\frac{u}{c}[/tex])
⇒f'/f =( 1 - [tex]\frac{u}{c}[/tex])
⇒(f - f')/f = [tex]\frac{u}{c}[/tex]
⇒∆f = f([tex]\frac{u}{c}[/tex])
so change in frequency, ∆f = f([tex]\frac{u}{c}[/tex])
now f = 410 × 10⁶ Hz , u = 525 m/s and c = 3 × 10 m/s
so, ∆f = (410 × 10⁶ × 525)/(3 × 10⁸)
= 71750 × 10⁻²
= 717.75 Hz ≈ 718 Hz
As a result, the fighter jet recorded a 718 Hz change in frequency.
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A person places a cup of coffee on the roof of her car while she dashes back into the house for a forgotten item. When she returns to the car, she hops in and takes off with the coffee cup still on the roof. A) If the coefficient of static friction between the coffee cup and the roof of the car is 0.20, what is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance. B) What is the smallest amount of time in which the person can accelerate the car from rest to 24 m/s and still keep the coffee cup on the roof?
(a) The maximum acceleration the car can have without causing the cup to slide is 1.96 m/s².
(b) The smallest amount of time in which the person can accelerate the car from rest to the final speed is 12.25 seconds.
What is the maximum acceleration of the car?
The maximum acceleration the car can have without causing the cup to slide is calculated as follows;
For the car not to slide, the applied force must equal the force of friction, so the that the acceleration of the car is zero.
F - Ff = ma
F - Ff = 0
F = Ff
ma = μmg
a = μg
where;
μ is the coefficient of static frictiong is acceleration due to gravitya = 0.2 x 9.8 m/s²
a = 1.96 m/s²
The smallest amount of time in which the person can accelerate the car from rest to the final speed is calculated as follows;
v = u + at
v = 0 + at
t = v/a
t = (24 m/s) / (1.96)
t = 12.25 s
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Two blocks (X and Y) are in contact on a horizontal frictionless surface. A 36-N constant force is applied to X as shown. The force exerted by X on Y is:
D 30 N F = ma yields 36N = 24kga, or a 1.5m/s2 acceleration. Due to the 4kg block's contact force, the 20kg block is accelerating. 20 kg at 1.5 m/s2 contact equals 30 N.
What is accelerating?
In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. They are accelerations and vector quantities.
The direction of an object's acceleration is determined by the direction of the net force acting on it.
Due to the 4kg block's contact force, the 20kg block is accelerating. 20 kg at 1.5 m/s2 contact equals 30 N.
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A cyndrical flask of cross sectional area A is fitted with an airtight piston that is free to slide up and down .contained within the flask is an ideal gas. Initially the pressure applied by the piston is 140kpa and the base of the flask is 22cm. When additional mass is added to the piston the pressure increases to 380kpa. Assuming the system is always at the temperature of 290k find the new height of the piston
Assuming the system is always at the temperature of 290k, the new height of the piston be 16.21 cm.
What is pressure?The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa).
Absolute, atmospheric, differential, and gauge pressures are different types of pressure.
As height of the piston is proportional to volume of ideal gas. From Boyle's law of gases at constant temperature we can write:
PV = pv
⇒PH = ph
⇒ 140 kpa × 22 cm = 290 kpa × h
⇒ h = 22 × 140/190 cm = 16.21 cm.
Hence, the new height of the piston be 16.21 cm.
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find the mass in [g] of the quantity of ice required at 0oc that 10 grams of 100oc steam will melt completely. the heat of vaporization of water at standard temperature and pressure is 2256 kj/kg. the heat of fusion of ice is 334 kj/kg. incorrect answer:
The mass of ice required at 0oc that 10 grams of 100oc steam will melt completely is 80.07 g.
The amount of energy that must be supplied to a solid substance (typically in the form of heat) in order to cause a change in its physical state and convert it to a liquid is known as latent heat of fusion (when the pressure of the environment is kept constant). The latent heat of fusion of one kilogramme of water, for example, is 333.55 kilojoules, which is the amount of heat energy that must be supplied to convert one kilogramme of ice without changing the temperature of the environment (which is kept at zero degrees Celsius).
Given,
mass of steam, [tex]m_s[/tex] = 10g = 0.01 kg
Latent heat of fusion of ice ,[tex]L_f= 334kj/kg[/tex]
Latent heat of vaporization of water, [tex]L_v=2256kj/kg[/tex]
change of temperature, [tex]\delta T=100-0=100\textdegree\ C[/tex]
The amount of heat cooling can be determined by.
[tex]m_{ice}L_f=m_sL_v+m_sS \delta T[/tex]
Here, S= specific heat constant i.e [tex]4.184 kj/kg[/tex]
[tex]m_{ice}L_f=(0.01*2256)+(0.01*4.184*100)\\\\m_{ice}*334=22.56+4.184\\\\m_{ice}*334=26.744\\\\m_{ice}=\frac{26.744}{334}=0.0807kg=80.07 g[/tex]
Thus, the mass of quantity of ice is 80.07 g.
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when light passes from a material with a high index of refraction into material with a low index of refraction: is the meaning of?
When light passes from a material with a high index of refraction into material with a low index of refraction, it is the meaning of a)incident angle. So, correct option is a.
Refraction of light is quite possibly of the most regularly noticed peculiarity, however different waves like sound waves and water waves additionally experience refraction. Refraction makes it feasible for us to have optical instruments, for example, amplifying glasses, focal points and crystals. It is likewise a direct result of the refraction of light that we can shine light on our retina.
The refractive file, additionally called the record of refraction, depicts how quick light goes through the material.
The refractive index is dimensionless in nature. For a given material, the refractive record is the proportion between the speed of light in a vacuum (c) and the speed of light in the medium (v). On the off chance that the refractive record for a medium is addressed by n, it is given by the accompanying recipe:
n=c/v
where n is refractive index, c is the speed of light in vacuum and v is the speed of light in other medium.
Hence, correct option is a.
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(Complete question) is:
when light passes from a material with a high index of refraction into material with a low index of refraction: it is the meaning of?
a)incident angle
b)speed of light
c)refraction angle
d)all of the above
figure 21-5b shows the concentration profile of a band of material after it has traveled through a column for 2 min and for 26 min. why is the band broader after 26 min? (b) explain why a chromatographic separation normally has an optimum flow rate that gives the best separation.
The broader band because of mobility of ions and because of peaks developed it has optimum flow rate.
There are two characteristics of the concentration profile that are important in determining the efficiency of a column and its ability to separate or separate subsequent solute zones. The first peak maximum refers to the position of maximum concentration in the peak. To obtain sufficient resolution, the maxima of two adjacent peaks must be separated from each other. Such separations depend on the nature of the solutes and the selectivity of the stationary and mobile phases.
A second characteristic that is important for efficiency and resolution is peak width. Peaks with distant maxima are still broad and may not fully resolve the solutes. For this reason, peak width is very important in chromatography.
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Light of wavelength 550 nm illuminates a double slit, and the interference pattern is observed on a screen behind the slit. The third maximum is measured to be 2.8 cm from the central maximum. The slits are then illuminated with light of wavelength 440 nm.
How far is the fourth maximum from the central maximum?
The fourth maximum is measured to be 2.97 cm from the central maximum when slits are illuminated with light of wavelength 440 nm.
The interference or bending of waves passing through an aperture around the corner of an obstacle or into the geometric shadow area of the obstacle/aperture is defined as diffraction. Secondary sources of propagating waves are diffractive objects or apertures. It is the process by which a ray or other wave system expands as it passes through a narrow aperture or edge. It is usually the result of interference between the generated waveforms.
Given,
[tex]\lambda = 550 nm=550*10^{-9}n[/tex]
[tex]y_3 = 2.8 cm = 2.8 * 10^{-2} m[/tex]
The third maximum from the central maximum can be determined by
[tex]y_3 = \frac{3D \lambda}{d}\\\\2.8*10^{-2}=\frac{3*550*10^{-9}D}{d}\\\\\frac{2.8*10^{-2}}{3*550*10^{-9}}=\frac{D}{d}\\\\0.0169*10^6=\frac{D}{d}[/tex]
Now, the fourth maximum from the central maximum can be determined by,
Now, [tex]\lamda = 440 nm =440 *10^{-91} m[/tex]
[tex]y_4 = \frac{4D \lambda }{d}\\\\y_4=4*0.0169*10^6*440*10^{-9}\\\\y_4=29.744*10^{-3}m=2.97 cm[/tex]
Thus, the fourth maximum is measured to be 2.97 cm from the central maximum
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when a digital fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the: input phosphor. output phosphor. photocathode. focusing lenses.
When a digital type of fluoroscopy unit's gets magnification button and used it, then the resulting magnification decrease in size due to input phosphor. Option (a) is correct.
The front phosphor (or enter phosphor) is a element of the photo intensifier in fluoroscopic structures that converts the strength from x-rays into mild photons. It consists of a fluorescent cloth including cesium iodide activated with sodium (CsI:Na) and coats the doorway floor of the photo intensifier.
Output phosphor consists of a fluorescent compound (referred to as P20) fabricated from silver-activated zinc cadmium sulphide (ZnCdS:Ag) particles. In addition, caesium iodide has a crystalline structure, and the enter phosphor may be synthetic in order that the slim needle-like crystals (about five μm in diameter) are laid down perpendicular to the screen.
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Correct Question:
when a digital fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the:
a). input phosphor.
b). output phosphor.
c). photocathode.
d). focusing lenses.
in the instant of fig. 11-41, two particles move in an xy plane. particle p1 has mass 6.5 kg and speed v1 2.2 m/s, and it is at dis- tance d1 1.5 m from point o. particle p2 has mass 3.1 kg and speed v2 3.6 m/s, and it is at distance d2 2.8 m from point o. what are the (a) magnitude and (b) direction of the
The magnitude and direction of net angular momentum of particles is [tex]52.7kg.m^2/s[/tex] out of the page.
Given,
particle 1:-
Mass ,m1 = 6.5 kg
speed , v1 = 2.2 m/s
distance from point O, d1 = 1.5 m
Particle 2:-
Mass ,m2 = 3.1 kg
speed, v2 = 3.6 m/s
distance from point O, d2 = 2.8 m
Angular momentum can be determined by formula,
p=mvd
angular momentum of particle-1, [tex]\rho_1=6.5*2.2*1.5=21.45 kg.m^2/s[/tex]
angular momentum of particle-2, [tex]\rho_2=3.1*3.6*2.8=31.248 kg.m^2/s[/tex]
Then, the net angular momentum is
[tex]\rho=\rho_1+\rho_2=21.45+31.248=52.69kg.m^2/s[/tex]
The direction of angular momentum of particle is always perpendicular to the motion of particle, as the particles are moving on xy-plane then the their direction will be perpendicular to xy-plane i.e. z-plane, it means out of the page.
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Your question is incomplete, here is the complete question.
In the instant shown in the diagram, two particles move in an xy-plane. Particle P1 has mass 6.5 kg and speed v1 = 2.2 m/s, and it is at distance d1 = 1.5 m from point O. Particle P2 has mass 3.1 kg and speed v2 = 3.6 m/s, and it is at distance d2 = 2.8 m from point O. What is the magnitude and direction of net angular momentum of the two particles about O?
A) 52.7 kg · m2/s out of the page B) 52.7 kg · m2/s into the page C) 21.5 kg · m2/s into the page D) 9.8 kg · m2/s into the page E) 9.8 kg · m2/s out of the page
8
Figure 2.24 shows a model of a skyscraper which Omar made
for the school exhibition. It is mounted on a firm plastic base
and has a total weight, W, of 4.0 N. Omar applies a force Fas
shown to test the stability of his model.
Calculate the minimum value of F required to cause the model
to rotate about point P.
Answer:
24-997
Explanation:
determine whether each geologic feature is being caused by tensional, compressional, or shear stresses by analyzing the directions of the forces being applied. drag the appropriate items to their respective bins. each item may be used only once.
Tensional faults—normal faults—cause each geologic feature, whereas shear faults—strike–slip faults—cause the remaining three.
What names do rocks above and below a fault go by?The hanging wall, also known as the headwall or footwall, is the higher or overlaying block along the fault plane as rocks slide past one another during faulting. The footwall is the lower block.
What error occurs when the footwall shifts uphill while the hanging wall shifts downward?A typical fault occurs when the rocks above the fault plane, also known as the hanging wall or footwall, move downward in relation to the rocks below the fault plane, or the footwall. A reverse fault occurs when the footwall slides up relative to the hanging wall.
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a lunar eclipse occurs when group of answer choices the earth lies directly between the sun and the moon the sun lies directly between the earth and the sun the moon lies directly between the sun and the earth
A lunar eclipse occurs when the earth is directly between the sun and the moon, thus, its shadow creates a lunar eclipse. Correct answer: letter A.
A lunar eclipse occurs when the Earth's shadow is cast on the surface of the Moon. This can only happen if the Sun, Earth, and Moon are all aligned in a straight line. In this configuration, the Earth is directly between the Sun and the Moon, blocking the light from the Sun and casting a shadow onto the Moon.
What is a lunar eclipse?Is an astronomical event that occurs when the Moon passes directly behind Earth and into its shadow. During a lunar eclipse, the Moon will appear dark as it is illuminated by only the faint, red-orange light of the sun scattered by Earth's atmosphere.
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radio frequencies (rf) can set off electronic initiating devices and some very sensitive explosives in the same manner as static electricity. electronic communication transmissions should not occur within a minimum of of suspected explosives.
Radio frequency (RF) is a unit of measurement used to describe the rate of oscillation of electromagnetic radiation spectrum, or electromagnetic radio waves, with frequencies as high as 300 gigahertz (GHz) and as low as 9 kilohertz (kHz).
What is RF in radio?The electromagnetic "spectrum" refers to all electromagnetic radiation in its collective form. Electromagnetic energy includes, for example, the radio waves and microwaves that transmitting antennas emit. "Radiofrequency" or "RF" energy or radiation is the term used to refer to all of them.The most often used frequencies are those that range from 500 MHz to 3 GHz because they provide a balanced set of benefits in terms of transmission range, capacity to handle large data rates, and antennae size. Most radar systems employ the X bands (8–12 GHz).Furthermore, you must have your radio receiver tuned to a specific frequency in order to pick up the information carried by these waves. It so happens that 2.4 GHz and 5 GHz are the WiFi frequencies. Your microwave's frequency and these waves are extremely comparable.To Learn more About Radio frequency refer To:
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A copper rod has a length of 60 cm at room temperature (220C). The coefficient of expansion of copper is 17 x 10-6 /Co, its specific heat is 386 J/(kg K), its melting point is 1356 K and its heat of fusion is 207 kJ/kg. The copper rod is heated so that its length increases to 60.3 cm a. What is the new temperature of the copper rod? T 316 b. If the mass of the rod is 0.52 kg, how much thermal energy was added to the rod? Q 5.9 x104 J c. If the mass of the rod is 0.52 kg, how much total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature? Qー ? x105 J
There are 3 questions here that need to be solved.
a. The new temperature of the copper rod is 316°C (rounded)
b. If the mass of the rod is 0.52kg, the thermal energy that was added to the rod is 5.9 * 10^4J
c. If the mass of the rod is 0.52kg, the total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature by using coefficient of thermal expansion of copper at 17 * 10^-6 / C° is 3.21 * 10^5J
What is coefficient of thermal expansion?It refers to the rate at which material expands with increase in temperature. More specifically, this coefficient is determined at constant pressure and without a phase change, such as the material is expected to still be in its solid or fluid form.
Now, moving to the question.
a. What is the new temperature of the copper rod?
L₂ = L₁ + L₁ * α * (t₂ - t₁)
Where
L2 is length of the rod that already increased
L1 is the original length of the rod
α is coefficient of expansion of copper
t2 is the new temperature
t1 is the original temperature
60.3cm = 60cm + 60cm * 17 * 10^-6 * (t₂ - 22°C)60.3
= 60 + 0.00102 * (t₂ - 22)0.3 = 0.00102 * (t₂ - 22)t₂ - 22
= 0.3 / 0.00102 t₂ - 22 = 294.12
t₂ = 316°C (rounded)
b. If the mass of the rod is 0.52kg, how much thermal energy was added to the rod?
Q = m * c * Δt
m = mass
c = coefficient
= 0.52 * 386 * (316 - 22)
= 59011.68
= 5.9 * 10^4 J (rounded)
c. If the mass of the rod is 0.52kg, how much total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature?
Q = m * c * Δt + m * L fusion
= m * c * (t melt - t₁) + m * L fusion
= 0.52 * 386 * (1356 - (22 + 273)) + 0.52 * 207 * 10^3Q
= 320603.92
Q = 3.21 * 10^5 J
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❧ What are the effects of Refraction of Light?
The following are effects of refraction of light: bending of light, change in wavelength of light, splitting of light rays.
What is Refraction?Refraction is the bending of light rays as it passes from one medium into another. Light rays refracts whenever it travels at an angle into a substance with a different refractive index. This change of direction is caused by a change in speed of the light rays. For example, when light travels from air into water, it slows down, causing it to continue to travel at a different angle or direction.
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Miguel (m=72) and Fernando (89 kg) board the bumper cars at the local carnival. Miguel is moving at a velocity of 4 m/s when he rear-ends Fernando who is at rest in his path. Fernando and his 125-kg car lunge forward at 2 m/s. Determine the post-collision velocity of Miguel and his 125-kg car.
Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.
Define collision.
A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.
The conservation of momentum law allows for the calculation of Miguel's post-collision speed.
Mass times velocity (p=mv), a vector quantity, equals momentum.
Before the collision, Miguel and Fernando's car had a momentum of 782 kg*m/s (m1=72 kg) (v1=4 m/s) + (m2=125 kg) (v2=0 m/s).
The momentum of the system following the collision is equivalent to (m1=72kg)(v1=?m/s) + (m2=125kg)(v2=2m/s) = 782 kg*m/s.
Miguel's comment velocity is -1.2 m/s, as determined by solving for v1. Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.
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a cube of mass m sliding without frictionat speed v0. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity - both speed and direction - after the collision?
The cube's velocity - both speed and direction - after the collision is [tex]v=\frac{v_0}{5}[/tex].
What is kinetic energy conservation?
Kinetic energy, which may be seen in the movement of an object, particle, or group of particles, is the energy of motion. Any moving item uses kinetic energy, such as a person walking, a baseball being thrown, a piece of food falling off a table, or a charged particle in an electric field.
From kinetic energy conservation:
[tex]\begin{gathered}\frac{m v_0^2}{2}=\frac{m v^2}{2}+\frac{I \omega^2}{2} \\I=\frac{M d^2}{12}=\frac{2 m d^2}{12}=\frac{m d^2}{6} \\\frac{m v_0^2}{2}=\frac{m v^2}{2}+\frac{m d^2 \omega^2}{12} \\v_0^2=v^2+\frac{d^2 \omega^2}{6} \Longrightarrow(1)\end{gathered}[/tex]
From angular momentum conservation:
[tex]\begin{gathered}\frac{m v d}{2}=I \omega \\\frac{m v_0 d}{2}=\frac{m d^2 \omega}{6} \\v_0=\frac{d \omega}{3} \\d^2 \omega^2=9 v_0^2 \Longrightarrow(2)\end{gathered}[/tex]
Substituting (2) into (1):
[tex]\begin{gathered}v^2=v_0^2-\frac{3 v_0^2}{2} \\v^2=-\frac{v_0^2}{2}\end{gathered}[/tex]
The correct answer is: [tex]v=\frac{v_0}{5}[/tex]
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Problem 2: Students of College Physics use an experimental set-up like the one shown, except, of course, the cut-away section to aid visualization is only in the drawing. A cylinder with a d=11.43cm inner diameter and an hmax=15.56cm height is placed on a heating plate. A weighted circular disc fits snugly into the cylinder, creating a gas-tight seal; the combined mass of the disc and the weights it supports is 19.07kg.50% Part (a) The disc initially rests on the bottom of the cylinder, but argon gas is introduced through a valve until the bottom of the disc is at a height h0 which is 41.9% of hmax. How much argon gas, in moles, is in the cylinder if it has a temperature of 22.1∘C? (Assume that the gas in the cylinder approximates an ideal gas.)50% Part (b) Suppose an additional mass of 5.56kg is placed on the disc, but at the same time, the heating plate is used to heat the gas in the cylinder. If the height of the disc above the cylinder bottom is now 87.8% of hmax, then what is the change in the temperature, in degrees Celsius, of the gas?
(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.
(b) The change in temperature of the gas is approximately 976.7°C.
Part (a):
Given data:
Inner diameter of the cylinder (d) = 11.43 cm = 0.1143 m
Height of the cylinder ([tex]h_{max[/tex]) = 15.56 cm = 0.1556 m
Height of the disc above the bottom of the cylinder ([tex]h_0[/tex]) = 41.9% of h_max = 0.419 * 0.1556 m
Combined mass of the disc and weights (m) = 19.07 kg
Temperature of the gas (T) = 22.1°C = 22.1 + 273.15 = 295.25 K
We need to find the amount of argon gas in moles in the cylinder.
Step 1: Calculate the volume of the gas in the cylinder.
[tex]\[ V_{\text{gas}} = \pi \left(\frac{d}{2}\right)^2 \times h_0 \]\\\\\ V_{\text{gas}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.419 \times 0.1556 \]\\\\\ V_{\text{gas}} \approx 0.00121 \, \text{m}^3 \][/tex]
Step 2: Calculate the number of moles of argon gas using the ideal gas law.
[tex]\[ PV = nRT \][/tex]
Where P is the pressure of the gas (we assume it is constant), V is the volume of the gas, n is the number of moles, R is the universal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
[tex]\[ n = \frac{PV}{RT} \]\\\\\ n = \frac{1 \times 0.00121}{8.314 \times 295.25} \]\\\\\ n \approx 4.09 \times 10^{-5} \, \text{moles} \][/tex]
Part (b):
Given data:
Additional mass placed on the disc ([tex]m_{extra[/tex]) = 5.56 kg
New height of the disc above the bottom of the cylinder (h_new) = 87.8% of h_max = 0.878 * 0.1556 m
We need to find the change in the temperature of the gas.
Step 1: Calculate the new volume of the gas in the cylinder.
[tex]\[ V_{\text{gas\_new}} = \pi \left(\frac{d}{2}\right)^2 \times h_{\text{new}} \]\\\\\ V_{\text{gas\_new}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.878 \times 0.1556 \]\\\\\ V_{\text{gas\_new}} \approx 0.00178 \, \text{m}^3 \][/tex]
Step 2: Use the ideal gas law again to find the new temperature of the gas.
[tex]\[ n_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{RT_{\text{new}}} \]\\\\\\ T_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{n_{\text{new}} \times R} \]\\\\\ T_{\text{new}} = \frac{1 \times 0.00178}{4.09 \times 10^{-5} \times 8.314} \]\\\\\ T_{\text{new}} \approx 1271.95 \, \text{K} \][/tex]
Change in temperature:
[tex]\[ \Delta T = T_{\text{new}} - T \]\\\\\ \Delta T = 1271.95 - 295.25 \]\\\\\ \Delta T \approx 976.7 \, \text{K} \][/tex]
Result:
(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.
(b) The change in temperature of the gas is approximately 976.7°C.
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A series of four direct shear tests were conducted on a dry sand. The initial size of the specimens were 50 mm by 50 mm (in horizontal plane) and 20 mm high. The following test results were reported. Test Number Vertical stress (kPa) 50 100 200 400 Shear Stress at Failure (kPa) 35.5 71 142 284 Plot the test results in normal stress-shear stress space, draw the best-fit Mohr-Coulomb failure envelope, and report the friction angle. [Note: assume that the cohesion intercept is zero.] Friction angle (degrees)
35.375 degrees is the friction angle. Four direct shear tests were performed for on a dry sand. The specimens were 50 mm by 50 mm in size at first.
Angle of internal friction, often known as friction angle, is a parameter indicating how well a unit of rock or soil can sustain a shear force. When failure simply results from a shearing stress, an angle (), measured between the normal force (N) and resultant force (R), is reached ( S ). A shear test is intended to exert stress on a test sample so that it slides out of alignment with the applied forces along a plane.
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the forearm in the figure(figure 1) accelerates a 3.2 kg ball at 7.5 m/s2 by means of the triceps muscle, as shown. a) Calculate the torque needed. Ignore the mass of the arm. Answer in m*Nb) Calculate the force that must be exerted by the triceps muscle. Answer in N
The calculated answer for the torque is 7.812 Nm and the force is 25.2 N
In the given question, we have to find the torque needed and the force exerted by the triceps muscle.
We know that, when there is a net torque on the system, it causes angular acceleration and the angular momentum is no more conserved.
Here let be the arm is massless.
The torque can be written as
τ=Iα
Where I is the moment of inertia.
α is the angular acceleration.
So,
τ=mR^2 α
τ=mRa
Here we have used the relationship between angular and tangential acceleration, which is: α=a/R.
The position vector is: R = 0.31
The mass is: m =3.6
The tangential acceleration is: a =7 m/s2
So, the torque will be,
τ=3.6*0.31*7
=7.812
So, the required torque is 7.812 Nm.
And we know that torque is: τ=F.R
So, force is F = τ/r
= 7.812/0.31
=25.2 N
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which of the following is not a type of electromagnetic wave with lower energy than visible light waves? microwaves| radar waves| the waves that carry AM or FM broadcasts| the sound waves coming from your transistor radio| the waves that carry television transmissions
The sound waves coming from your transistor radio is not a type of electromagnetic wave with lower energy than visible light waves.
What is electromagnetic wave explain?
Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.
How do electromagnetic waves travel?
Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through the air and solid materials but also through the vacuum of space.
What are the 7 types of electromagnetic waves?
In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.
Thus, sound waves coming from your transistor radio are the correct option.
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Problem Statement: Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs, and what that motion is, when the coefficient of static friction between all surfaces is (a) js = 0.40, (b) ps = 0.50. Figure: с P 100 mm B 8 kg -200 mm 25° AO 8 kg
In comparison to protons and neutrons, electrons have a very low mass (9.10938356 10 31 kg). An electron weighs about 1837 times as much as a proton, with a mass of (1.67262191027kg). Atoms' nuclei almost entirely account for their mass. There are only protons and neutrons present inside the nucleus.
In physics, what is insignificant?
Negligible refers to the ability to ignore or neglect. There are occasions when a force in physics is so little that its impact on the overall phenomenon is negligible. When you throw an apple up, for instance, the apple draws the earth, yet the earth does not move because the force the apple exerts is very little or inconsequential.
Why is mass so small?
If object A weighs 1 gram and object B weighs 1 kilogram, then object A is said to have a negligible mass in comparison to object B because it weighs so little. In mathematical problems, we typically neglect trivial values to avoid adding complexity.
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A 75 kg person stands on a metric bathroom scale that reads in Newtons while in an elevator. The
elevator accelerates upward at 3 m/s/s. Determine the person's actual weight and the reading of the
scale under these conditions.
Answer: 735 N (actual), 960 N (accelerating up)
Explanation:
Actual weight = W = mg = (75 kg)(9.8 m/s²) = 735 N
Scale reading = N (normal force acting on the person by the elevator) = mg + ma = 735 N + (75 kg)(3.0 m/s²) = 735 N + 225 N = 960 N
That's why you feel heavier when you are accelerating up in an elevator
earth's atmosphere is most transparent to all wavelengths of light. earth's atmosphere is most transparent to all wavelengths of light. infrared visible x-ray ultraviolet both visible and ultraviolet
Earth's atmosphere is transparent for waves like visible light and microwaves is correct statement.
A wave is a planned, systematic transfer of disturbances from one place to another. In addition to the waves that move across the water's surface, which are the most well-known waves, there are also waves in sound, light, and the motion of subatomic particles. In the simplest waves, the disturbance oscillates repeatedly at a fixed frequency and wavelength (see periodic motion). Mechanical waves, like sound, need a medium to move, in contrast to electromagnetic waves, which can do so in a vacuum and do not need one (see electromagnetic radiation). How a wave moves through a medium depends on its properties.
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Each item below belongs either with the population of disk stars or the population of halostars of the Milky Way Galaxy. Match each item to the appropriate population.Disk stars:-high-mass stars-youngest stars-stars that all orbit in nearly the same plane-the SunHalo stars:-globular clusters-oldest stars-stars whose orbits can be inclined at any angle-stars with the smallest abundance of heavy elements
Oldest: Red Giant in globular cluster M13 AND red main-sequence star in globular cluster M13
Middle: the sun
Youngest: Hot, blue main-sequence star in disk
What is globular cluster?
Globular clusters, which include tens of thousands to millions of stars, are stable, firmly bound clusters. They are connected to all different kinds of galaxies. Globular clusters are strongly gravitationally bound and often larger than open clusters.
As a result, the Milky Way's globular clusters (which house extremely old stars) show a spherical component, whilst the open clusters, other young stars, and star-forming areas show a disk-shaped component.
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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/ 15. The lasers produce separate interference patterns on a screen a distance 4.40 m away from the slits. Part What is the distance Aymamin between the second maximum of laser 1 and the third minimum of laser 2 on the same side of the central maximum? Express your answer in meters. View Available Hint(s) Ay mas ma 5867 k m Submit Previous Answers x incorrect: Try Again; 3 attempts remaining
the distance Aymamin between the second maximum of laser 1 and the third minimum of laser 2 on the same side of the central maximum is 0.4 m
For part be we can use the equation
ym = m*λ*L/d
so for the first max for each one m =1
so for laser 1
y = (d/20)*(6.0m)/d
y = 6.0m /20 = 0.3
for laser 2 we do the same thing
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
(1) hence first maxima of laser1 is closest to central maxima
Δy = 0.4m -0 .3 m = 0.1m
For part C)
we use the same equation to find the y for laser 1, except m =2
y= 2*6.0 m / 20 = 0.6 m
Now for laser 2 we use:
ym=(m+1/2)*λ*L/d
since there is no central minimum the first minimum is at m = 0.that means that the third minimum is at m = 2
simplifying the equation we get
y = (2.5)*6.0 / 15 = 1m
now we solve
Δy= 1m - 0.6m = 0.4 m
Double Slit Experiment:
The double-slit experiment shows that light and matter can exhibit properties of both classically defined waves and particles. Furthermore, it demonstrates the fundamental stochastic nature of quantum mechanical phenomena.
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Ultraviolet radiation has frequencies from 3.0×1015 to 3.0×1016 Hz, whereas the frequency region for gamma ray radiation is 3.0×1019 to 3.0×1024 Hz.
We can say that:
1. The speed of ultraviolet radiation is _________ higher than/ lower than/the same as gamma ray radiation.
2. The wavelength of ultraviolet radiation is ________ longer than /shorter than/ the same as gamma ray radiation.
What is the maximum number of orbitals that can be identified with the following quantum numbers: n = 3, ℓ = 0
a. 7
b. 8
c. 18
d. 1
e. 2
1) The same as gamma rays. As we know all electromagnetic radiations travel with speed of light.
2) longer than gamma rays
As we know, higher the frequency lower is the wavelength.as ultraviolet has lower frequency than gamma rays.so it has longer wavelength.
3) n=3 ,l=0 ,hence it is 3s orbital so it has 1 orbital.
(d) is correct.
The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.
By comparing the values of the fields you provide, a straightforward frequency analysis generates a report that shows each value for those fields along with the frequency at which it appears.
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When forces are balanced they?
Answer:
There is no change in motion it is constant
Explanation:
Balanced forces are equal in size and opposite in direction. When forces are balanced, there is no change in motion. In one of your situations in the last section, you pushed or pulled on an object from opposite directions but with the same force.
An elevator, suspended by a cable, is moving upward at constant speed. The correct relationship between the magnitude of force of gravity on the elevator and tension force in the cable is: O A) the magnitude of force of gravity equals the magnitude of tension force B) the magnitude of force of gravity is greater than the magnitude of tension force C) the magnitude of force of gravity is less than the magnitude of tension force D) the magnitude of force of gravity is less than or equal to the magnitude of tension force E) the magnitude of force of gravity is equal to or greater than the magnitude of tension force
An elevator is traveling steadily upward while being hung by a cable. When this happens, the tension force's strength is equal to the gravitational force's strength.
The elevator, which is hung by a cable and traveling upward at a steady pace, is principally being acted upon by two forces. One of the forces is gravitational force, which is the weight of gravity acting on the elevator. Another force is tension in the cable, which is holding the elevator in place. Acceleration is the same as the velocity's time rate of change.
Where v is the end velocity and u is the beginning velocity, an is equal to v-u/t. However, the speed here is constant. Thus, there will be no acceleration. As a result, the net force acting on this elevator will also become zero utilizing Newton's Second Law of Motion.
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