Answer:
ThermodynamicsQuantum mechanicsNuclear physicsMechanicsAstrophysicsAnswer:
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A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting
Answer:
the person is sitting 1.5 m from the left end of the board
Explanation:
Given the data in the question;
Wb = 125 N
Wm = 500 N
T₂ = 250 N
Now, we know that;
T₁ + T₂ = Wb + Wm
T₁ + 250 = 125 + 500
T₁ = 125 + 500 - 250
T₁ = 375 N
so tension of the left chain is 375 N.
Now, taking torque about the left end
500 × d + 125 × 2 = 250 × 4
500d + 250 = 1000
500d = 1000 - 250
500d = 750
d = 750 / 500
d = 1.5 m
Therefore, the person is sitting 1.5 m from the left end of the board.
According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.
a. 1+ charge and 1- charge seperated by 200pm
b. 1+ charge and 1+ charge seperated by 100pm
c. 1+ charge and 1- charge seperated by 100pm
d. 2+ charge and 1- charge seperated by 100pm
According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.
Highest potential energy to lowest potential energy.
b. 1+ charge and 1+ charge seperated by 100 pm
a. 1+ charge and 1- charge seperated by 200 pm
c. 1+ charge and 1- charge seperated by 100 pm
d. 2+ charge and 1- charge seperated by 100 pm
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
Example 9.1
The Archer
Let us consider the situation proposed at the beginning of
this section. 160kg archer stands at rest on frictionless ice
and fires a 0.50-kg arrow horizontally at 50 m s (Fig. 9.2).
With what velocity does the archer move across the ice after
firing the arrow
v1f = -0.16 ms
Explanation:
Use the conservation law of linear momentum:
m1v1i + m2v2i = m1v1f + m2v2f
where
v1i = v2i = 0
m1 = 160 kg
m2 = 0.50 kg
v2f = 50m/s
v1f = ?
So we have
0 = (160 kg)v1f + (0.5 kg)(50 m/s)
v1f = -(25 kg-m/s)/(160 kg)
= -0.16 m/s
Note: the negative sign means that its direction is opposite that of the arrow.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Answer:
[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]
Explanation:
Given that,
The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]
A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :
[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]
Put all the values,
[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]
So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a) S = 63.2 km/h
b) V = 63.2 km/h*(-0.316 , 0.949)
Explanation:
Let's define:
North as the positive y-axis
East as the positive x-axis.
Also, remember the relation:
Distance = Time*Speed
Let's assume that she starts at the position (0km, 0km)
Then she travels due North at 90km/h for two hours, then the displacement is
90km/h*2h = 180km to the north
Then the new position is:
(0km, 180km)
Then she travels West at 60km/h for one hour.
Then the distance traveled to the West (negative x-axis) is:
60km/h*1h = 60km to the west
Then the new position is:
(-60km, 180km).
a) The average speed is defined as the quotient between the displacement and the time.
We know that the total time traveled is 3 hours.
And the displacement is the difference between the final position and the initial position.
this is:
D = √( -60km - 0km)^2 + (180km - 0km)^2)=
D = √( (60km)^2 + (180km)^2) = 189.7 km
Then the average speed is:
S = (189.7 km)/(3 h) = 63.2 km/h
b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.
So, if the final position is (-60km, 180km)
We need to find a vector that represents the same angle, but that is on the unit circle.
Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:
(-60km/ 189.7 km, 180km/ 189.7 km)
(-60/189.7 , 180/189.7)
We can just check that the module of the above versor is 1.
[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]
Then the average velocity is:
V = 63.2 km/h*(-60/189.7 , 180/189.7)
We can simplify our versor so the velocity equation is easier to read:
V = 63.2 km/h*(-0.316 , 0.949)
for the long life cells we have to connect them in ____ combination
Answer:
Parallel combination.
for the long life cells we have to connect them in parallel combination
hope it is helpful to you
Computer use ___code to transmit information
Binary code is the answer
Answer:
binary code is the answer of blank
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equilibrium point and released. Determine
a) the spring constant
b) the maximum velocity of the mass
c) the maximum acceleration of the mass
d) the total mechanical energy of the mass
e) the period and frequency of the mass and spring and
f) the equation of time-dependent vertical position of the mass
Answer:
a) [tex]k=19.6N/m[/tex]
b) [tex]V_m=0.81m/s[/tex]
c) [tex]a_m=6.561m/s^2[/tex]
d) [tex]K.E=0.096J[/tex]
e) [tex]T=0.78sec[/tex] &[tex]F=1.29sec[/tex]
f) [tex]mx'' + kx' =0[/tex]
Explanation:
From the question we are told that:
Stretch Length [tex]L=0.150m[/tex]
Mass [tex]m=0.30kg[/tex]
Total stretch length[tex]L_t=0.150+0.100=>0.25[/tex]
a)
Generally the equation for Force F on the spring is mathematically given by
[tex]F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}[/tex]
[tex]k=19.6N/m[/tex]
b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by
[tex]V_m=A\omega[/tex]
Where
A=Amplitude
[tex]A=0.100m[/tex]
And
[tex]\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s[/tex]
Therefore
[tex]V_m=A\omega\\\\V_m=8.1*0.1[/tex]
[tex]V_m=0.81m/s[/tex]
c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by
[tex]a_m=\omega^2A[/tex]
[tex]a_m=8.1^2*0.1[/tex]
[tex]a_m=6.561m/s^2[/tex]
d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}*0.3*0.8^2[/tex]
[tex]K.E=0.096J[/tex]
e)
Generally the equation for the period T is mathematically given by
[tex]\omega=\frac{2\pi}{T}[/tex]
[tex]T=\frac{2*3.142}{8.1}[/tex]
[tex]T=0.78sec[/tex]
Generally the equation for the Frequency is mathematically given by
[tex]F=\frac{1}{T}[/tex]
[tex]F=1.29sec[/tex]
f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by
[tex]mx'' + kx' =0[/tex]
Where
'= signify order of differentiation
Consider a sample containing 1.70 mol of an ideal diatomic gas.
(a) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(b) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
(c) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(d) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
I don't know
because I don't know
What is 11 divided by 73370 I need to know how you found the answer
You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take the positive -direction to be upward, and choose y 0 to be the point where the stone leaves your hand. Find the stone's position 1.50s after it leaves your hand.
Express your answer with the appropriate units.
Find the y-component of the stone's velocity 1.50 s after it leaves your hand. Express your answer with t0he appropriate units.
Answer:
The velocity after 1.5 s is 22.7 m/s downwards.
Explanation:
Initial velocity = - 8 m/s
acceleration, a = - 9.8 m/s2
time, t = 1.5 s
Use first equation of motion
v = u + at
v = - 8 - 9.8 x 1.5
v = - 8 - 14.7
v = - 22.7 m/s
Thus, the velocity after 1.5 s is 22.7 m/s downwards.
Why is it that, when we observe an extragalactic source whose diameter is about one lightday, we are unlikely to see fluctuations in light output in times shorter than about one day
The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.
How to observe extragalactic sources whose diameter is about one light day?When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.
For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.
In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.
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The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum
From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.
Answer:
time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.
Explanation:
Height, h = 15 m
Newton's second law
Force = mass x acceleration
The unit of gravitational force is Newton and the value is m x g.
where, m is the mas and g is the acceleration due to gravity.
Let the time of fall is t.
Use second equation of motion
[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]
Let the final speed is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]
how can the starch be removed from the leaves of potted plants
Answer:
Explanation:
There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.
En 2.0 s, una particula con aceleración constante a lo largo del eje x se mueve desde x =10 m
hasta x =50 m. La rapidez al final del recorrido es de 10 m/s. ¿Cuál es la aceleración de la partícula?
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
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Mention & Instrument used to measure
the mass of the body.
Answer:
a scale is used to measure the mass of the body
You want to calculate how long it takes a ball to fall to the ground from a
height of 20 m. Which equation can you use to calculate the time? (Assume
no air resistance.)
O A. vz? = v? +2aAd
B. a =
V₂-vi
At
O c. At=V1
4
a
O D. At=
2Ad
a
If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:
D. Δt= √2Δd/a
What is the equation for finding the length of time for a free fall?The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².
So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.
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PLEASE HELP!! URGENTT!
Answer:
120 Newton
Explanation:
Given the following data;
Mass = 12 kg
Angle = 4°
We know that acceleration due to gravity is equal to 10 m/s
To find the minimum force to stop the block from sliding;
Force = mgCos(d)
Where;
m is the mass of an object.
g is the acceleration due to gravity.
d is the angle of inclination (theta).
Substituting into the formula we have;
F = 12*10*Cos(4°)
F = 120 * 0.9976
F = 119.71 ≈ 120 Newton
A balloon pops, making a loud noise that startles you. What kind of energy best describes this experience?
A. Thermal Energy
B. Sound Energy
C. Gravitational Energy
D. Radiant Energy
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the following distances from the axis of the rod, where distances are measured perpendicular to the rod's axis.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]
E = 1461.95 N/C
c) The electric field E is calculated as:
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]
E = 239.76 N/C
3. Four charges having charge q are placed at the corners of a square with sides of length L. What is the magnitude of the force acting on any of the charges
Answer:
Fr = 1.91 * 9*10⁹*q²/L²
Explanation:
Let´s say that the corners of the square are A B C and D
We are going to find out the force on the charge placed on B ( the charge placed in the upper right corner.
As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.
Force due to charge placed in A
module Fₓ = K* q² / L² in the direction of x
Force due to charge placed in C
module Fy = K* q²/L² in the direction of y
Force due to the charge placed in D
That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.
d² = L² + L² = 2*L²
d = √2 * L
The module of the force due to charge place in D
F₄₅ = K*q²/ 2*L²
To get the force we need to add first Fₓ and Fy
Fx + Fy = F₁
module of F₁ = √ Fx² + Fy² the direction will be the same as the diagonal of the square then:
F₁ = √ ( K* q²/L² )² + ( K* q²/L² )²
F₁ = √ 2 * K*q²/L²
And now we add forces F₁ and F₄₅ to get the net force Fr on charge in point B.
The direction of Fr is the direction of the diagonal and is of rejection
the module is
Fr = F₁ * F₄₅
Fr = √ 2 * K*q²/L² + K*q²/ 2*L²
Fr = ( √ 2 + 0,5 ) * K*q² /L²
K = 9*10⁹ Nm²C²
Fr = 1.91 * 9*10⁹*q²/L²
We don´t know units of L and q
How can i prove the conservation of mechanical energy?
Answer:
We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved
Explanation:
who is corazon aquino?
Answer:
Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.
Answer:
Former President of the Philippines
Explanation:
define emperical formula and what is the dimensional formula of force and energy
Answer:
An empirical formula represents the simplest whole number ratio of various atoms present in a compound.The dimensional formula of force is [[tex]MLT^{-2}[/tex]]The dimensional formula of energy is [[tex]ML^{2} T^{-2}[/tex]]Darwin believed that emotional expressions began as ________ that came to have evolutionary value because they ________. Select one: a. physiological reactions; increased the efficiency of bodily reactions b. communication devices; increased the efficiency of bodily reactions Incorrect c. physiological reactions; convey emotional states to other members of the species d. random mutations; convey expectations to other members of the species
Answer:
c. physiological reactions; convey emotional states to other members of the species
Explanation:
Darwin believed that emotional expressions began as physiological reactions that came to have evolutionary value because they convey emotional states to other members of the species. These reactions are used by many different types of species to convey various things.
During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during this trial
Answer: hello your question is poorly written attached below is the complete question
answer :
TA = 1.6*10^-24 * 60 * 2, TB = 1.6*10^-24 * ( 60 + 30 ) * 2 -- ( option 1 )
Explanation:
a = 2m/s^2
Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц
Tb - Ta = m₂ a
∴ Tb = m₂ a + Ta
= ( 30 * 1.6 * 10^-24 * 2 ) + ( 60 * 1.6 * 10^-24 * 2 )
= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц
Can anyone help
Me please the question is on the photo that I attached it to
Answer:
2.8 MW
Explanation:
There are 7 wind turbines in the wind farm as shown in the diagram. Thus, the energy output by one turbine is 1/7 if the total energy output. So, 19.6/7=2.8MW
The liquid and gaseous state of hydrogen are in thermal equilibrium at 20.3 K. Even though it is on the point of condensation, model the gas as ideal and determine the most probable speed of the molecules (in m/s). What If? At what temperature (in K) would an atom of xenon in a canister of xenon gas have the same most probable speed as the hydrogen in thermal equilibrium at 20.3 K?
Answer:
a) the most probable speed of the molecules is 409.2 m/s
b) required temperature of xenon is 1322 K
Explanation:
Given the data in the question;
a)
Maximum probable speed of hydrogen molecule (H₂)
[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )
where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K
molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol
we substitute
[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³ )
[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³ )
[tex]V_{H_2[/tex] = 409.2 m/s
Therefore, the most probable speed of the molecules is 409.2 m/s
b)
Temperature of xenon = ?
Temperature of hydrogen = 20.3 K
we know that;
T = (Vxe² × Mxe) / 2R
molar mass of xenon; Mxe = 131.292 g/mol
so we substitute
T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314 )
T = 21984.14167 / 16.628
T = 1322 K
Therefore, required temperature of xenon is 1322 K
