(a) The average kinetic energy of a molecule of an ideal gas is 6.21 x 10⁻²¹ J.
(b) The total random kinetic energy of the molecule in 1 mole of this gas is 3,741.3 J.
(c) The RMS speed of the oxygen molecule is 215.25 m/s.
Average kinetic energy of ideal gasThe average kinetic energy of a molecule of an ideal gas is calculated as follows;
[tex]K .E = \frac{3}{2} K T[/tex]
Where;
K is Boltzmann constant = 1.38066 x 10⁻²³ J/KT is temperature in Kelvin, = 27°c + 273 = 300 K[tex]K.E = \frac{3}{2} \times (1.38 066 \times 10^{-23}) \times 300\\\\K.E = 6.21 \times 10^{-21} \ J[/tex]
Total random kinetic energyThe total random kinetic energy of the molecule in 1 mole of this gas is calculated as follows;
[tex]K.E = \frac{3}{2} nRT\\\\K.E = \frac{3}{2} (1) (8.314)(300)\\\\K.E = 3,741.3 \ J[/tex]
RMS of speed of oxygen moleculeThe RMS speed of the oxygen molecule is calculated as follows;
[tex]K.E = \frac{1}{2} mV_{rms}^2\\\\V_{rms}^2 = \frac{2K.E}{m} \\\\V_{rms} = \sqrt{\frac{2 K.E}{m} }[/tex]
one mole of oxygen gas = 32 g = 0.032 kg
[tex]V_{rms} = \sqrt{\frac{2 \times 741.3 }{0.032} }\\\\V_{rms} = 215.25 \ m/s[/tex]
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What was made first? The color orange or the food orange? it's not on my test I just want to know.
Answer:
the fruit was made before the color
Answer:
I think the color because I lowkey think the food Orange was named after the color bc ofc it was Orange.
Explanation:
Please help me!!! I'm so lost!!
Robert throws a 3 kg rock at 20 m/s. What is the rock's momentum?
a
O 0.15 kgm/s
O 6.67 kgm/s
O 20 m/s
O 60 kg . m/s
An atom has 5 protons, 4 neutrons, and 6 electrons. Which one of the following could be its isotope
Answer:
9 :)
Explanation:
Question 1: What is the before image showing?
Question 2: What is the after image showing?
Answer:
the after image is showing precipitation. i dont know what the before image is though.
Explanation:
4. Brian stands at the edge of a cliff and throws a stone horizontally over the edge with a
speed of 18 m/s. The cliff is 50 m above a flat, horizontal beach: (2 pts each)
a. How long does it take for the stone to reach the bottom of the cliff?
b. What is the range of the stone?
5. A student shot a basketball at an initial velocity of 25 m/s at an angle of 30 degrees with
respect to the horizontal: (2 pts eac)
a. What is the vertical component of the initial velocity?
b. What is the maximum height reached by the arrow?
6. A fireman stands 50 m away from a burning building and directs a stream of water from
ground level at an angle of 30° above the horizontal. If the speed of the stream as it
leaves the hose is 40 m/s: (2 pts each)
a. How long will it take for the stream to reach the building?
b. At what height will the water strike the building?
Answer:A).3.16 s B).56.92
A).12.5m/s
B).7.81m
Explanation:
A toy car has a mass of 3 Kg is pushed horizontally with a force of 3 N. Friction is so small it can be ignored.
What is the acceleration?
How fast is it going after 8 s if it starts from rest?
How far did it go?
Explanation:
Mass = 3kgForce=3NAcceleration= Force/mass=3/3=1 m/s²v=u+atv=0+1(8) (At rest)v=8m/sDisplacement = velocity× timeDisplacement=8×8 mDisplacement=64mGravity is a force that acts upon a ball after it is thrown
true
false
Answer:
true.
Explanation:
The forces on a thrown ball after it leaves the thrower's hand are the force of gravity .
fluid meaning in Nepali
Answer:
Tarala
Explanation:
It's Nepali
Answer:
Fluid meaning in Nepali is Tarala (तरल )
Example Compute the capacitance of a parallel-plate capacitor with plates 10cm by 20cm, separation of 5mm, and a mica dielectric.
Answer:
Explanation:
Mica dielectric constant (k) is approximately 3. The distance between the plates is 5 mm = [tex]5\times10^{-3}[/tex] m = d and the area of each plate is[tex]A = 0.1\times 0.2=2\times 10^{-2} m^{2}[/tex]. So, the capacitance of that capacitor is:
[tex]C=\frac{\epsilon_{0}kA}{d}=\frac{(8.85\times10^{-12})(3)(2\times10^{-2}}{5\times10^{-3}}=10.62\times10^{-11} =1.062\times10^{-10} F[/tex]
or approximately 0.106 nF.
a ball is rolled at a velocity of 12 miles per second. after 36 seconds, it comes to a stop. what is the acceleration of the ball?
Question :-
A Ball is Rolled at a Velocity of 12 m/s. After 36 sec , it comes to a stop. What is the Acceleration of the ball ?Answer :-
Acceleration is -0.33 m/s² .Explanation :-
As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .
For calculating the Acceleration , we will use the Formula :-
[tex] \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} [/tex]
Where ,
V denotes to the Final VelocityU denotes to the Initial VelocityT denotes to the Time TakenTherefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } [/tex]
[tex] \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} [/tex]
[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}[/tex]
[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}[/tex]
[tex] \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} [/tex]
Hence :-
Acceleration of Ball is -0.33 m/s² .[tex] \underline {\rule {212pt} {4pt}} [/tex]
explain why it is dangerous to jump from a fast moving train
Answer:
When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.
Explanation:
you be in serious injury.
The blood pressure of human body is greater at the feet than at the brain. why?give reason
Answer:
Gravity stronger closer to the core of the earth
Explanation:
The gravity applied to the feet is stronger than to the upper part of the body for example the brain because of the distance it is between the body part and the core of the earth where the gravity force is pulling towards. Even though the difference between the gravity force between the brain and the feet is minimal it is still a greater force at the feet than at the brain
Can u please explain the force of gravity in this
Answer:
gravity is a natural phenomenon in which objects with mass,energy,light are attracted to one another
A circular ferris wheel that revolves at a constant rate once every 30 seconds. The
radius of the ferris wheel is 10 m. What is the normal force of the ferris wheel on a 10
kg toddler at the very bottom of the ferris wheel?
The normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
Centripetal force of the Ferris wheelThe normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is calculated as follows;
Fn = Fc + mg
Fn = mω²r + mg
Fn = m(ω²r + g)
where;
ω is the angular speed = 1 rev/30 s = 2π/30 s = 0.21 rad/sFn = 10(0.21² x 10 + 9.8)
Fn = 102.41 N
Thus, the normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
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A gold wire carries a current of 76.2 mA.
(a) Find the number of electrons that flow past a given point in the wire in 10.4 minutes.
[Answer] electrons
b)In what direction do the electrons travel with respect to the current?
o opposite direction
o same direction
o The magnitude is zero.
#a
Time=10.4min=10.4(60s)=624sCurrent=76.2mA=76.2×10^{-3}A=INow
Charge=Q[tex]\\ \tt\Rrightarrow Q=\dfrac{I}{t}[/tex]
[tex]\\ \tt\Rrightarrow Q=\dfrac{76.2\times 10^{-3}}{624}[/tex]
[tex]\\ \tt\Rrightarrow Q=0.122\times 10^{-3}[/tex]
[tex]\\ \tt\Rrightarrow Q=122\times 10^{-6}C[/tex]
[tex]\\ \tt\Rrightarrow Q=122\mu C[/tex]
#b
Same directionHow do you write numbers in
scientific notation? Explain
with an example.
Answer:
3.34x10*5
Explanation:
this is just a example.
take 3 sig.fig
Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and comes to a complete stop .24 seconds after impact.
Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
What happens when one side of the tug of war rope has a large net force than the other
Answer:
When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.
Explanation:
When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.
A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS = 700 W/m2 , oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is αS = 0.83, and the efficiency of conversion of the absorbed flux to electrical power is η = P/αSGSA = 0.553 − 0.001 K−1 Tp, where Tp is the panel temperature expressed in kelvins and A is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which Tsur = T[infinity] = 35°C, h = 10 W/m2 ⋅K, and (b) a breezy winter day, for which Tsur = T[infinity] = −15°C, h = 30 W/m2 ⋅K. The panel emissivity is ε = 0.90
Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0[/tex]
[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s[/tex]
(a) the electrical power generated for still summer day
[tex]T_s = T_{\infty} = 35 ^oC = 308 \ k[/tex]
[tex][0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k[/tex]
[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W[/tex]
(b)the electrical power generated for a breezy winter day
[tex]T_s = T_{\infty} = -15 ^oC = 258 \ k[/tex]
[tex][0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k[/tex]
[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W[/tex]
A spring has a relaxed length of 7 cm and a stiffness of 200 N/m. How much work must you do to change its length from 10 cm to 15 cm
The work W needed to stretch/compress a spring from rest by a distance x is
W = 1/2 kx²
where k is the spring constant.
This means the work needed to change the length of this spring by 10 cm = 0.01 m is
W = 1/2 (200 N/m) (0.01 m)² = 0.01 J
and by 15 cm = 0.015 m is
W' = 1/2 (200 N/m) (0.015 m)² = 0.0225 J
Then the total work performed on the spring by stretching from 10 cm to 15 cm is
∆W = W' - W = 0.0225 J - 0.01 J = 0.0125 J
Which of the following statements are true about the motion of an object?
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
b. A force perpendicular to the momentum changes the direction of the momentum but not its magnitude.
c. When an object moves at constant speed along a curving path, the net force on the object must act straight outward from the center of the kissing circle.
d. To make an object turn to the left, something has to exert a force to the right on the object.
Answer: B
Explanation: A force perpendicular to the momentum changes the direction of the momentum but not its magnitude.
Answer:
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
d. To make an object turn to the left, something has to exert a force to the right on the object.
Explanation:
For motion along a circle or curve, the net force on the particle must be parallel with the change in momentum.
This change in momentum of the system is given by;
[tex]\delta P_{sys} = F_{net} \delta t[/tex]
Thus, option A is correct.
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
Also, applying the principle of torque, to make an object turn to the left, something has to exert a force to the right on the object.
Option D is also correct.
A block-spring system oscillates on a frictionless horizontal surface. The time needed for the block to complete one cycle is 0.1 sec. Determine the time needed for the block to travel from -A/2 to A/2, where A is the amplitude of motion.
The time required for the block to travel or complete it's half-cycle is 0.05s
Data;
time = 0.1sdistance = -A/2 to A/2Amplitude of a Simple Harmonic MotionThis is the maximum displacement of a body from it's mean position.
Assuming a system start from its mean position i.e mean initial position and go to A/2 towards right and then back to his mean position after going through -A/2 then it is said to have completed one cycle. If the block oscillate only between -A/2 to A/2 then it will complete half cycle then it's time period will be 0.05 sec.
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A car is uniformly accelerated at a rate of 2 m/sec2 for 12 sec. If the original speed of the car is 36 m/sec, what is its final speed?
Answer:
60m/s
Explanation:
v=u+at
v=36+(2×12)
v=36+24
v=60m/s
Does The stretched string of an archer’s bow has kinetic energy in it.
Answer:
The ball has potential energy just like the Longbow has when it is stretched. The other aspect of energy that the bow portrays is kinetic energy. Kinetic energy is the energy of motion. When the string is released, the potential energy is converted to kinetic energy in movement of the arrow
Explanation:
Answer:
It has potential energy not kinetic
Explanation:
A ball rolls 12m in 2.0s. What is the ball’s average velocity?
Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
the tip of the nail is sharp.why?
Answer:
To minimize area which in turn increases pressure
a 1000 kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at 140 m/ s takes how long to stop under full braking?
Answer:
0.21 SECONDS
Explanation:
THE CAR TRAVELS AT A SPEED OF 140M/S AND A DISTANCE AS THE PREVIOUS = 3M THEREFOR S=D/T, T=D/S= 3/140= 0.12 SEC
Why are force fields necessary to describe gravity?
A. Gravity is a type of air resistance.
B. Gravity acts between any two objects with charges.
C. Gravity is a noncontact force.
D. Gravity can only push objects apart.
Answer:
C
Explanation:
Gravity pulls objects towards earth's core without needing any contact with objects
if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?
Answer:
if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?
ANSWER 256n
The mass of the astronaut on the moon as compared to the earth will be [tex]M_m=97.85\ lg[/tex]
What will be the mass?The mass of any substance or body is defined as how much quantity of matter is present.
Now it is given in the question:
Weight of the astronaut on earth [tex]W_E=981\ N[/tex]
Weight of the astronaut on earth [tex]W_M=160\ N[/tex]
The mass of the astronaut on the moon will be calculated as:
Weight on the moon will be given as:
[tex]W_M=M_M\times g_m[/tex]
Weight on the earth will be given as:
[tex]W_E=M_E\times g_e[/tex]
The ratio of the gravity of the earth to the moon is given as
[tex]\dfrac{g_e}{g_m} =\dfrac{9.81}{1.62} =6[/tex]
The mass of the earth will be calculated as
[tex]W_E=M_E\times ge[/tex]
[tex]M_E=\dfrac{981}{9.81} =100\ kg[/tex]
Now taking the ratio of the weight of the earth to the moon :
[tex]\dfrac{W_E}{W_M} =\dfrac{M_E\times g_e}{M_M\times g_m}[/tex]
[tex]M_M= \dfrac{W_M\times M_E}{W_E} \times \dfrac{g_e}{g_m}[/tex]
Now by putting the value in the formula we get;
[tex]M_M=\dfrac{160\times 100}{981} \times 6=97.85\ kg[/tex]
Thus the mass of the astronaut on the moon as compared to the earth will be [tex]M_m=97.85\ lg[/tex]
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