The mole fraction of H2 is 0.509, the mole fraction of N2 is 0.408, and the mole fraction of Ar is 0.084.
To calculate the mole fraction of each gas, we need to use the following formula:
mole fraction of gas = moles of gas / total moles of gas
To find the moles of each gas, we need to use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We are given the pressure of each gas in torr, so we need to convert it to atm by dividing by 760 torr/atm. We can assume that the volume and temperature are constant for all the gases.
Calculations:
For H2 gas:
n(H2) = (406.0 torr / 760 torr/atm) * V / (0.0821 L*atm/mol*K * 298 K)
n(H2) = 0.0176 mol
For N2 gas:
n(N2) = (325.1 torr / 760 torr/atm) * V / (0.0821 L*atm/mol*K * 298 K)
n(N2) = 0.0141 mol
For Ar gas:
n(Ar) = (66.3 torr / 760 torr/atm) * V / (0.0821 L*atm/mol*K * 298 K)
n(Ar) = 0.0029 mol
The total moles of gas are:
n(total) = n(H2) + n(N2) + n(Ar)
n(total) = 0.0176 mol + 0.0141 mol + 0.0029 mol
n(total) = 0.0346 mol
Now we can calculate the mole fraction of each gas:
X(H2) = n(H2) / n(total) = 0.0176 mol / 0.0346 mol = 0.509
X(N2) = n(N2) / n(total) = 0.0141 mol / 0.0346 mol = 0.408
X(Ar) = n(Ar) / n(total) = 0.0029 mol / 0.0346 mol = 0.084
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Potentiometric measurements are based on galvanic electrochemical cells. these types of cells consist of:_______
Two-electron or metallic conducting electrodes that are connected by an electrolyte solution that conducts ions.
What are potentiometric measurements?An electrochemical measurement method is potentiometry. The phrase was first used in relation to possible electrochemical measurement chain determinations. Potentiometer is a tool that measures an electrochemical cell's potential without drawing current or changing the makeup of the cell.
Potentiometric measurement is done to detect the difference in potential between a working (an indicator) electrode and a counter (a reference), using a potentiometer. The working/indicator electrode is called the cathode (right half-cell) and the counter/reference electrode is called an anode(half of a left cell).
A basic galvanic cell may just have one electrolyte isolated from it by a semi-permeable membrane, but a more complex one has two independent half-cells linked together by a salt bridge. An inert electrolyte like potassium sulfate is included in the salt bridge, and its ions will diffuse into each of the individual half-cells to balance the charges that are building up at the electrodes.
I understand the question you are looking for is this:
Potentiometric measurements are based on galvanic electrochemical cells. These types of cells consist of:
two inert metal electrodes to which an electrical potential is applied.two-electron or metallic conducting electrodes that are connected by an electrolyte solution that conducts ions.a biological recognition element and a physicochemical transducer, often an electrochemical or optical device.a polarizable working electrode to which an external voltage is applied with the resulting cathodic or anodic current of the cell being monitored.Learn more about galvanic electrochemical cells here:
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When operating an aircraft at cabin pressure altitudes above 12,500 feet msl up to and including 14,000 feet msl, supplemental oxygen shall be used during?
Supplemental oxygen shall be used in that flight time in excess of 30 minutes at those altitudes.
Most plane cabins are pressurized to 8,000 feet above sea level, an altitude that lowers the amount of oxygen inside the blood by approximately 4 percentage points, researchers say.
The higher the altitude, the less oxygen there is in the air and the lower the overall air pressure is. If flights were not pressurized, passengers could be prone to various physiological illnesses. Because of this, federal policies require that all commercial flights over 8,000 feet be pressurized.
The flight crew must use supplemental oxygen for the entire duration of flight operations above a cabin pressure altitude of 14,000 feet MSL.
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A proton exits the cyclotron 1. 0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1. 0 ms ?.
The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.
The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.
The kinetic energy for the protons can be computed by using the formula:
K.E = 1/2mv²mv² = 2 K.E
v = sqrt( 2*KE / M)
the kinetic energy of the medical isotopes = 6.5 MeV
substituting the values,
v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )
v = 3.53 × 10⁷ m/s
The radius of the orbit can be estimated by using the formula:
mv² / R = qvB
r = q*v / mv²
r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )
r = 0.19415 m
Since diameter (D) = 2r,
D= 2(0.19415 m)
D= 0.39 m
D≅ 39 cm
The time period to complete a revolution around the spiral trajectory is:
T = 2πr / v
T = 2*3.14* 0.1941 / 3.53*10^7
T = 0.7 × 10⁻⁷ s
Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:
n = t / T
n = 10^-3 / (0.7*10^-7)
n = 14285.71
n ≅ 14000 revolutions
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Your question is incomplete, but most probably full question was:
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.
(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous
(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?
What force is large enough to change the direction of a planet?
A. Gravity
B. Friction
C.Tension
D. Air Resistance
Answer:
Gravity
Explanation:
The more massive the object is, the higher the gravity
Hence, a massive star, like our sun, has enough gravity to pull on planets and keeping them in orbit
what is kinematics ?
explain ~
[tex] \\ \\ [/tex]
tysm! :)
Answer:
Kinematics is the study of the motion of mechanical points, bodies and systems without consideration of their associated physical properties and the forces acting on them. The study is often referred to as the geometry of motion, and it models these motions mathematically using algebra
Answer:
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion is kinematics
A mountain skier has begun descending the 30o slope. if the coefficient of kinetic friction is 0.10, what is her acceleration?
The acceleration of the body is found to be -4.05 m/s^2
What is the acceleration?We define the acceleration as the rate of change of the velocity with time. Hence, we can write that;
a = (μcosθ - sinθ)g
a = acceleration
μ = coefficient of kinetic friction
θ = angle of inclination
g = acceleration due to gravity
Thus;
a = (0.10cos 30 - sin 30) 9.8
a = -4.05 m/s^2
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An engine has an input of heat energy of 10,750 J and does 2,420 J of work. Which of the following is the heat loss?
Group of answer choices
a. 13,200 J
b. 8,330 J
c. 4.44 J
c. 0.225 J
B. The heat loss by the heat engine is 8,330 J.
Heat loss by the heat engine
The heat loss by the heat engine is calculated as follows;
H = E - W
where;
H is heat lossE is input energyW is work doneH = 10,750 J - 2,420 J
H = 8,330 J
Thus, the heat loss by the heat engine is 8,330 J.
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How much power will be required to force a current of 4.13 amps to flow through a conductor whose resistance is 113 ohms? Use two decimals for your answer. Round your answer to two decimals.
The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts
What is power?This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:
Power (P) = square current (I²)× resistancet (R)
P = I²R
How to determine the powerCurrent (I) = 4.13 AResistance (R) = 113 ohmsPower (P) =?P = I²R
P = 4.13² × 113
P = 1927.43 watts
Thus, the power required is 1927.43 watts
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The image above shows a crash test dummie's head traveling at -1.005 m/s while striking a headrest from a car traveling 4.524 m/s with a mass of 2005.6 kg. If the crash test dummie's head head bounces off the headrest with a 9.965 m/s and the car continues traveling at 4.487 m/s, calculate the mass of the crash test dummie's head?
Answer:
From the calculation, the mass of the dummies head is 1647.44 Kg
What is the mass of the crash test dummies head?We know that the momentum after collision is equal to the momentum before collision.
Mass of the headrest = 2005.6 kg
Initial velocity of the head rest = 4.524 m/s
Final velocity of the head rest = 4.487 m/s
Mass of the dummy = m
Initial velocity of the dummy = -1.005 m/s
Final velocity of the dummy = 9.965 m/s
Then;
(m * -1.005) + (2005.6 * 4.524 ) = (2005.6 * 4.487) + (m * 9.965)
-1.005m + 9073.33 = 8999.13 + 9.965m
9073.33 - 8999.13 = 9.965m + 1.005m
18072.46 = 10.97m
m = 18072.46/ 10.97
m = 1647.44 Kg
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A person riding a Ferris wheel moves through positions at (1) the top, (2) the bottom, and (3) mid-
height. If the wheel rotates at a constant rate, rank these three positions according to
(a)the magnitude of the person’s centripetal acceleration,
(b) the magnitude of the net centripetal force on the person, and
(c) the magnitude of the normal force on the person from the bottom, and
(d) the weight the person feels, greatest first.
A person riding a Ferris wheel moves through positions at
(1) the top,
(2) the bottom, and
(3) mid-height. If the wheel rotates at a constant rate, rank these three positions according to
(a)the magnitude of the person’s centripetal acceleration, is same in all cases.
(b) the magnitude of the net centripetal force on the person, is same in all three cases.
(c) the magnitude of the normal force on the person from the bottom,
2<3<1
(d) the weight the person feels, greatest first is at the bottom.
What is the relation between force and acceleration?The centripetal acceleration [tex]a_{c}[/tex] exists provided by : a= [tex]v^{2} /r[/tex] , therefore the magnitude depends upon the speed and the radius. This represents that in the case of the Ferris wheel the centripetal acceleration includes a constant magnitude so its value exists the same in all cases.From Newton's second law the force is provided by F=ma and since a exists constant in magnitude , F is also constant in magnitude. and its value exists the same in all three cases.At the top [tex]a_{c}[/tex] stands directed downward so the normal force exist given by [tex]F_{N}[/tex]=[tex]mg[/tex]- [tex]ma_{c}[/tex]While at the bottom [tex]a_{c}[/tex] exist directed upward and the normal force is provided by [tex]F_{N}[/tex]=mg +m[tex]a_{c}[/tex]Meanwhile,at mid -height [tex]a_{c}[/tex] stands directed horizontally to the inside so it doesn't have a vertical component so its normal force is provided by [tex]F_{N}[/tex]=mgThe normal force exists greatest at the bottom than at mid-height than at the top.To learn more about centripetal force , refer :
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Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They collide, couple, and roll away together with a combined momentum of
The combined momentum of the passengers is 5000 kgm/s.
Combined momentum of the passengerThe combined momentum of the passengers is calculated as follows;
P = mv1 + mv2
where;
m is mass of the passengersv1 is velocity of the first passengerv2 is velocity of the second passengerP = m(v1 + v2)
P = 5000(-1 + 2)
P = 5000 kgm/s
Thus, the combined momentum of the passengers is 5000 kgm/s.
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A velocity-time graph is shown below:
The average acceleration in the first 10 seconds of the journey is 2 m/s²; option B.
What is acceleration?Acceleration is the rate of change of velocity with time.
Acceleration = change in velocity/timeFrom the graph;
Acceleration in first seconds = 20 - 0/5 = 4 m/s²
Acceleration in second 5 seconds = 0
Average acceleration = 4 + 0/2
Average acceleration = 2 m/s²
In conclusion, average acceleration is the average of the final and initial acceleration.
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The pressure due to the atmosphere is very large, why are we not crushed?
Answer:
It is because of the blood pressure balances with our atmospheric pressure
The length of a rectangular plot of land is 151.28 m, and its width is 72.35 m. What is the perimeter of this plot?
605.1 m
223.6 m
289.4 m
a.
b. 447.3 m
c.
d.
Please select the best answer from the choices provided
OA
OB
OC
OD
[tex] \large{ \boxed{ \tt{447.3 \: m}}}[/tex]
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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction acting on the box is a constant 20 N. How much work is done by the 50 N force?
The work is done by the 50 N force is 200 J.
Work done by the 50 N forceWork done = Fd
where;
F is Force applied d is the displacement of the objectWork done = 50 N x 4 m = 200 J
Thus, the work is done by the 50 N force is 200 J.
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Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coefficient of kinetic friction between the player and the ground is 0.70. His speed at the start of the slide is 8.23 m/s. I a) Calculate his acceleration during the slide. b) How long (in time) does he slide until he stops?
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
How to determine the force of frictionMass (m) = 81.5 KgCoefficient of friction (μ) = 0.7Acceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 NFrictional force (F) =?F = μN
F = 0.7 × 798.7
F = 559.09 N
A. How to determine the accelerationMass (m) = 81.5 KgFrictional force (F) = 559.09 NAcceleration (a) =?a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
B. How to determine the time Initial velocity (u) = 8.23 m/sFinal velocity (v) = 0 m/sDecceleration (a) = -6.86 m/s²Time (t) =?a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
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Josie ran at an average speed of 16 m/s. if her mass 10 kg, what was her kinetic energy as she crosses her finish line?
Answer: 1280 J
[tex]KE=\frac{1}{2}mv^{2} \\\\KE= \frac{1}{2} (10 kg)(16 m/s)^2\\=1280 J[/tex]
I have to be honest with you . . . It's not possible to determine the answer with only the information given in the question.
Kinetic energy = (1/2) (mass) (speed squared).
Josie's Kinetic energy as she crosses the finish line is
(1/2) (her mass) (her speed as she crosses the finish line)².
That's (5 kg) x (her speed as she crosses the finish line)² .
But read the question again.
WE DON'T know her speed as she crosses the finish line.
The question only tells us her average speed for the whole race.
We can be confident that her Average Kinetic Energy was
(5 kg) (16 m/s)² = 1,280 Joules.
But we can't tell what her kinetic energy was as she crossed the finish line, beause we don't know what her speed was at that instant.
Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 m to 3200 m below the surface. What is the gauge pressure at a 2452-m deep vent, assuming that the density of water does not vary
If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.
Calculation:
Step-1:
It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.
The gauge pressure at a particular depth of ocean water is calculated as:
[tex]$$P=\rho g h$$[/tex]
Here [tex]\rho[/tex] is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.
Step-2:
Now we are substituting the values to calculate the pressure at the depth of 2452-m.
[tex]$$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$$[/tex]
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A kettle transfers 6,000j of energy electrically. 1,500j of this is wasted. what is the efficiency of this kettle?
Answer:75 percent
Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,
output energy=input energy - wasted energy
output energy=6000j-1500j
output energy=4500
put that into the formula
efficiency =output energy/input enrgy x100%
efficiency=4500/6000 multiplied by 100%
efficiency=0.75x100%
efficiency=75%
1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.
Taking into account the definition of calorimetry, the final temperature of the silver and water is 11.84 °C.
CalorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m. c is the specific heat substance. ΔT is the temperature variation.Final temperature of the silver and waterIn this case, you know:
For silver:Mass of silver= 500 g= 0.5 kg (being 1000 g= 1 kg)Initial temperature of silver= 250 °CFinal temperature of silver= ?Specific heat of silver= 240 [tex]\frac{J}{kgC}[/tex] For water:Mass of water = 1000 g= 1 kgInitial temperature of water= 5 ºCFinal temperature of water= ?Specific heat of water = 4180 [tex]\frac{J}{kgC}[/tex]Replacing in the expression to calculate heat exchanges:
For silver: Qsilver= 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature of silver - 250 C)
For water: Qwater= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature of water - 5 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the silver gives up will be equal to the heat that the water receives. Therefore:
- Qsilver = + Qwater
And the final temperature of the silver is equal to the temperature of the water (Final temperature of silver= Final temperature of water= Final temperature). Then:
- 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature - 250 C)= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature - 5 C)
Solving:
- 120 [tex]\frac{J}{C}[/tex]× (Final temperature - 250 C)= 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]× (250 C - Final temperature) = 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]×250 C - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 4180 [tex]\frac{J}{C}[/tex]×5 C
30,000 J - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 20,900 J
50,900 J= 4300 [tex]\frac{J}{C}[/tex]×Final temperature
50,900 J÷ 4300 [tex]\frac{J}{C}[/tex]= Final temperature
11.84 °C= Final temperature
Finally, the final temperature of the silver and water is 11.84 °C.
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in the photoelectric effect a photon with an energy of 5.3 * 10^-19 J strikes an electron in a metal
The velocity of the photo electron is 6.11 × 105 ms
Given :
Supplied energy, Es = 5.3 x 10-19 J
Minimum energy of the electron to escape from the metal, E. = 3.6 x 10-19 J
To Find :
Velocity of photoelectron
Solution : The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So
5.3 × 10^-19 J = 3.6 × 10^-19 J + K
K = 5.3 x 10^-19 - 3.6 x 10^-19
K = 1.7 × 10^-19 J
The formula of kinetic energy is given by:
K = 1/2 mv^2
v = √2K/m
= √2 x 1.7 x 10^-19
√ 9.1 x 10^-31
v = 6.11 x 10^5 m/s
So, the velocity of the photo electron is 6.11 x 10^5 m/s
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question is down below
The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s
What are the components of velocity?We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.
Hence;
Vertical component = 40 m/s sin 15 degrees = 10.35 m/s
Horizontal component = 40 cos 15 degrees = 38.6 m/s
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How is a rainbow formed?
Answer:
easy
Explanation:
it when the sunlight hits a rain droplets ,some of the light is reflected .The electromagnetic spectrum is made of light with many different wavelengths and each is reflected at a different angle . Thus spectrum is seperated producing a rainbow
As the electric potential near the ground increases during a thunderstorm, a positive charge current can move up pointed objects, such as masts of ships, producing a luminous halo or glow known as ____.
Answer:
St. Elmo's Fire - well known to old time sailors
What will be the pressure exerted by the object if 5,000 N of force
area of 200 cm²?
Answer:
[tex]250000 \frac{n}{m {}^{2} } [/tex]
Explanation:
pressure is the force applied perpendicular to the surface area of an object per unit area.
for the question,
Force (F) = 5000N
[tex]Area = 200cm {}^{2} = 0.02m {}^{2} \\ Pressure = \frac{force}{area} \\ Pressure = \frac{5000}{0.02} \\ Pressure = 250000 \frac{n}{m {}^{2} } [/tex]
A 15.2 kg mass has a gravitational potential energy of -342 J. How high from the ground is it? Group of answer choices GPE cannot be negative 2.3 m 22.5 m 530 m
A. The height above the ground reached by the mass is 2.3 m.
Height traveled by the mass
P.E = mgh
where;
m is mass h is height reached by the objectg is acceleration due to gravityh = P.E/mg
h = (342)/(15.2 x 9.8)
h = 2.3 m
Thus, the height above the ground reached by the mass is 2.3 m.
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20. Q: How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
A: 1.56 s
B: 2.04 s
C: 2.45 s
D: 3.72 s
Answer:
2.45s
Explanation:
is explanation needed too?
a pulley is assumed massless and frictionless and rotates freely about its axle. the blocks have masses m1=40g and a block m2= 20g, and block m1 is pulled to the right by a horizontal force magnitude f=0.03n. find the magnitude of acceleration of a block m2 and the tension in the cord.....?
The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.
Given:
mass of block 1, m₁ = 40 gm = 40×10⁻³ kg
mass of block 2, m₂ = 20 gm = 20×10⁻³ kg
Applied force, F = 0.03 N
Calculation:
Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:
F = ma
where F is the applied force
m is the total mass of the system
a is the acceleration of block 2 (as it is pulled by horizontal force)
From the above equation we get:
0.03 N = (m₁+m₂) a
⇒ a = (0.03 N) / (m₁+m₂)
⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)
⇒ a = 0.5 m/s²
Now, from the free-body diagram of block 2 as shown in figure 3, we get:
Balancing the forces along the horizontal:
∑Fₓ = 0
∴ T = m₂ a
where T is tension in the string
a is the acceleration of block 2
Applying values in the above equation we get:
T = (20×10⁻³ kg) × (0.5 m/s²)
= 0.01 N
Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.
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Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) Determine the Earth’s average orbital speed expressed in kilometers per hours.
b) Based on the information given in this question, calculate the approximate mass of the Sun.
The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg
Relationship between Linear and angular speedLinear speed is the product of angular speed and the maximum displacement of the particle. That is,
V = Wr
Where
V = Linear speedW = Angular speedr = RadiusGiven that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed
W = 2[tex]\pi[/tex]/T
W = (2 x 3.143) / (365.26 x 24)
W = 6.283 / 876624
W = 7.2 x [tex]10^{-4}[/tex] Rad/hr
The Earth’s average orbital speed V = Wr
V = 7.2 x [tex]10^{-4}[/tex] x 149.6 x [tex]10^{6}[/tex]
V = 107225.5 kilometers per hours.
b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law
M = (4[tex]\pi ^{2}[/tex][tex]r^{3}[/tex]) / G[tex]T^{2}[/tex]
M = (4 x 9.8696 x 3.35 x [tex]10^{24}[/tex]) / (6.67 x [tex]10^{-11}[/tex] x 7.68 x [tex]10^{11}[/tex])
M = 1.32 x [tex]10^{26}[/tex] / 51.226
M = 2.58 x [tex]10^{24}[/tex] Kg
Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg
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