The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
The given parameters;
Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, [tex]\lambda_{rms}[/tex] = 4.32 x 10⁻⁸ mMean collision time, [tex]\tau = 3 \times 10^{-10} \ s[/tex]The mean velocity of the gas molecules is calculated as follows;
[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]
The temperature of the gas molecules is calculated as follows;
[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]
where;
k is Boltzmann constant[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]
The number of gas molecules per unit volume is calculated as follows;
[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]
The pressure of the gas molecule is calculated as follows;
[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]
Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
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Suppose you want to make the demonstration more dramatic by attracting as much hair as possible with the balloon. Which of the following is the best choice to accomplish your goal? A. Decrease the electric force by holding the balloon closer to the hair. B. Increase the electric force by rubbing the balloon for a longer period of time. C. Decrease the electric force by using a larger balloon. D. Increase the electric force by holding the balloon farther away from the hair.
Answer:
B. Increase the electric force by rubbing the balloon for a longer period of time.
Explanation:
5. Layer of Earth consisting of crust & upper layer of mantle ________
Answer:
lithosphere
Explanation:
hope this helps you!!
A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this
Answer:
fifth harmonic
Explanation:
objects want to ______ ___________ doing what they're __________ ____________ because they are "lazy." This is called __________.
Answer:
Explanation:
Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.
The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 34 N. Assume the incline is smooth and determine the angle of the incline if the block moves with constant speed.
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]
[tex]\Rightarrow mg\sin{\theta} = F[/tex]
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get
[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]
or
[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]
[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]
[tex]\;\;\;=42.9°[/tex]
Someone help me please !!!! Will mark Brianliest !!!!!!!!!!!!!!!!
Answer:
Decant it.
Explanation:
Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.
Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.
explain how water erosion changes land forms
Answer:
Water erosion can slowly wash away dirt and rocks, wearing away landforms and sometimes forming rivers that can even carve out stone over time.
a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
Answer:
when the ball is at its highest point
Explanation:
Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.
In what way does Isaac represent us?
Answer:
I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us
Explanation:
A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?
4 m/s
36 m/s
30 m/s
1.44 m/s
4. If a charged body is negatively charged, it has excess of _______
Answer:
Electrons
Explanation:
Electrons are negatively charged sub-atomic particles, therefore when a body's negatively charged, it means that there's more electrons than protons.
Explain how the removal of heat energy affects the speed of the particles in a substance
Answer:
The removal of heat energy slows the speed of particles
Explanation:
When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.
A crate is released on a frictionless plank inclined at angle q with
respect to the horizontal. Which of the following relationships is true?
(Assume that the x-axis is parallel to the surface of the incline.)
a. Fy = Fg
c. Fy = Fx
b. Fx = 0
d. none of the above
None of the given options is correct based on the relationship for the components of the crate's weight.
The given parameter:
Angle of inclination, = qThe vertical component of the force on the crate is calculated as follows;
[tex]F_y = W \times cos(q)\\\\F_y = F_gcos(q)[/tex]
The horizontal component of the crate is calculated as follows;
[tex]F_x = F_g \times sin(q)\\\\F_x = F_g sin(q)[/tex]
Thus, we can conclude that none of the given options is correct based on the relationship for the components of the crate's weight.
Learn more about forces on inclined weight here: https://brainly.com/question/11351170
Write down the condition required for the thermonuclear fusion.
Answer:
The main condition necessary for a controlled thermonuclear fusion:
The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.
A 2.2 kg model rocket is shot straight up in the air from the ground, with an initial velocity of 36.4 m/s. The rocket reaches its maximum height, and falls back to the ground. What is the maximum height of the rocket? Round your answer to 2 decimal places.
Answer:
Explanation:
Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.
PE = KE
mgh = ½mv²
h = v²/2g
h = 36.4²/ (2(9.81))
h = 67.53109...
h = 67.53 m
A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) what is the mass of the block office? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?
Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]
Subsequently,
[tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
[tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 [tex]m \ s^{-1}[/tex].
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
i need help with the problem below
Answer:
Explanation:
a) F = ma
a = F/m
a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s
b) t = v/a
t = 200 / 7.2 x 10⁻⁶
t = 2.8 x 10⁷ s about 10½ months
c) v² = u² + 2as
s = (v² - u²) / 2a
s = (200² - 0²) / (2( 7.2 x 10⁻⁶))
s = 2.8 x 10⁹ m nearly 7 times around the earth
And all this assumes NO FRICTION.
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
plz help ASAP I'll mark as brainliest
Hi there!
1.
Hooke's law states that:
F = -kx
k = Spring constant (N/m)
x = DISPLACEMENT from equilibrium (m)
Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.
2.
The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)
3.
The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.
4.
We can begin by looking at the given graph.
When the spring force = 4N, the total length of the spring is 35 cm.
Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:
35 - 30 = 5 cm.
5.1.
If the spring elongates by 10 cm, the total length of the spring is:
30 + 10 = 40 cm
According to the graph, a length of 40 cm corresponds to a force of 8N.
5.2.
We can solve for the weight of the ball using the following:
W (weight) = m (mass) · acceleration due to gravity (10N/kg)
Using a summation of forces:
∑F = T - W
The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:
0 = T - W
T = W = 8 N
5.3.
0 = T - Mg
T = Mg
Use the prior value of T and gravity to solve:
8 = 10M
m = 0.8 kg
What is the force exerted on a selectively permeable membrane because water has moved from an area of higher concentration to an area of lower concentration of water?
Diffusion
Facilitated transport
Osmotic pressure
Endocytosis
Question
what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?
Answer
OSMOSIS
Explanation
Osmosis: This is the hydrostatic force acting to equalize the concentration of water on both sides of the membrane that is impermeable to substances dissolved in that water. Water will move along its concentration gradient.
HEY!!^^
have a good day^^
thank me later^^^
carryonlearing^^
stuck with this one.
A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:
a- 18.3
b- 12.0
c- 11.1
d- 14.4
Answer:
D.
correct me if im wrong
brainlest plsss<333
You are given two vectors A⃗ =−3.00i^+5.00j^ and B⃗ =7.00i^+2.00j^. Let counterclockwise angles be positive.
a)What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
Express your answer in degrees.
b)What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
c)Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees.
Answer:
Explanation:
We can subtract directly the corresponding components and check using the parallelogram rule.
Explanation:
Have a look:
enter image source here
Where, graphically, I used the fact that:
→
A
−
→
B
=
→
A
+
(
−
→
B
)
For the magnitude we use Pythagoras (with the components) to get:
∣
∣
∣
→
A
−
→
B
∣
∣
∣
=
√
(
−
1
)
2
+
(
5
)
2
=
√
1
+
25
=
√
26
≈
5.1
For the direction I can see that will be
90
∘
from the
x
axis up to the
y
axis, plus the little bit passed the
y
axis given as:
θ
=
arctan
(
1
5
)
=
11.3
∘
giving in total: angle
=
90
∘
+
11.3
∘
=
101.3
∘
Using the formula below, calculate the kinetic energy of the 6 gram stone going 10 Mph
going 10 mph
KE = 1/2 Mass x Speed?
=
Answer:
Explanation:
A wonderful (NOT) mixed unit problem
convert mph to m/s
v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)
v = 4.4704 m/s
KE = ½(0.006)4.4704²
KE = 0.05995342848 J
KE = 0.06 J
The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure Q2(b). Determine: i) the velocity of the marble at B ii) the horizontal distance R of the basket from the end of the tube, and iii) the speed at which the marble falls into the basket.
Crazy Wally Ok Ok ok hhahahaha
25.Figure 22.22 shows a plot
of potential versus position
along the x-axis. Make a plot
of the x-component of the
electric field for this situation.
The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.
dV = - E . ds
E = [tex]- \frac{dV}{ds} \ \hat s[/tex]
Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.
Let's apply this expression for each section of the given graph:
1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.
The derivative of a constant is zero.
E = 0
2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.
We look for the equation of the line.
V-V₀ = m (x- x₀)
We carry out the derivative.
E = - m i ^
The slope (m) is:
[tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]
Let's calculate.
[tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]
Let's substitute.
E = [tex]2 \hat i \ V/m[/tex]
3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.
We look for the equation of the line and we derive.
E = - m i ^
Let's substitute.
[tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]
E = - 4 [tex]\hat i[/tex] V / m
4) From x₀ = 4.5 m to x_f = 6m. The potential is constant and the derivative of a constant is zero.
E = 0
5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]
E = - 0.5 [tex]\hat i[/tex] V/m
6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]
Let's substitute.
E = 2 [tex]\hat i[/tex] V/m
7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]
Let's substitute.
E = 1 [tex]\hat i[/tex] V/m
In the attachment we can see these Electric fields as a function of distance.
In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
Learn more about the electric field here: brainly.com/question/14306881
an object is traveling with a constant velocity of 5m/s. How far will it have gone after 7s
Answer:
35m/s
Explanation:
Answer:
35m/s
Explanation:
Simply multiply 5 and 7.
What is the significance of Isaac's name?
Answer:
ghgivjgifigo ra together aigig disgust u hoodie
Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton
hits the ball back at a velocity of -44 m/s. Determine the impulse Josh
Hamilton's bat delivered to the baseball and enter to the nearest kg-m/s with
the correct sign.
The Impulse delivered to the baseball is 89 kgm/s.
To solve the problem above, we use the formula of impulse.
⇒ Formula:
I = m(v-u)................. Equation 1Where:
I = Impulse delivered to the baseballm = mass of the baseballv = Final velocity of the baseballu = initial speed of the baseballFrom the question,
⇒ Given:
m = 0.8 kgu = 67 m/sv = -44 m/s⇒ Substitute these values into equation 1
I = 0.8(-44-67)I = 0.8(-111)I = -88.8I ≈ -89 kgm/sNote: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.
Hence, The Impulse delivered to the baseball is 89 kgm/s.
Learn more about impulse here: https://brainly.com/question/7973509
What is the half-life of the imaginary element Lokium? Show your work
Answer:What is the half life of the element Lokium?
The half-live of the element Lokium is 4.
Explanation:
sorry if im wrong, have good day
Answer:
The half-life of the element Lokium is 4
What is half-life?
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
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