Answer:
1. 10³ m²
2. 32.4 ml
3. 1.91 × 10²¹ molecules
Explanation:
Here is the complete question
1. Estimate the size of a one-molecule-thick oil film formed by spreading 1 ml of oil on the surface of the water. Assume that an oil molecule is roughly 10 nm in size. Show your work.
2. If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.
3. Estimate the number of molecules in 1 ml of oil assuming they’re 10 nm in size. What assumptions do you have to make? Show your work.
Solution
1. Since the film would cover an area, A, and would have a height which is the thickness of the molecule, h = 10 nm = 1 × 10⁻⁹ m, its volume is V = Ah. This volume also equals the volume of the oil. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, V = 1 × 10⁻⁶ m³.
The size of the oil drop is its area. So, A = V/h
= 1 × 10⁻⁶ m³ ÷ 1 × 10⁻⁹ m
= 10³ m²
2. The area of the tank is 10 in by 10 in = 100 in². Since we want to cover half the area, we require 100 in²/2 = 50 in² = 50 in² × (0.0254)² m²/in² = 0.0324 m².
If the thickness of oil is one molecule thick which is 10nm = 1 × 10⁻⁹ m, the volume of oil is then thickness × area = 1 × 10⁻⁹ m × 0.0324 m²
= 0.0324 × 10⁻⁹ m³
= 32.4 × 10⁻⁶ m³
= 32.4 ml since 1 × 10⁻⁶ m³ = 1 ml
3. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, we need to find the volume of one molecule. Since it is assumed to be a sphere, its volume is V' = πd³/6 where d = size of oil molecule = 10 nm = 1 × 10⁻⁹ m.
Let n be the number of molecules present in 1 ml, then nV' = 1 ml = 1 × 10⁻⁶ m³. So, n = 1 × 10⁻⁶ m³/V' = 1 × 10⁻⁶ m³ ÷ πd³/6 = 6 × 10⁻⁶ m³/πd³
Substituting d into the equation, we have
n = 6 × 10⁻⁶ m³/π(1 × 10⁻⁹ m)³
n = 6 × 10⁻⁶ m³/π × 10⁻²⁷ m³
n = 1.91 × 10²¹ molecules
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Answer:8pi
Explanat:Omega =2pi/T
A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
What is the dimensional formula of time (T)?
Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
Answer :
[T] = [W(F)^-1(V)^-1]
If distance between two charges increased by 2 times then force
Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
In an experiment to measure the value of pi, the following results are obtained for pi:
3.14 3.11 3.20 3.06 3.08
a) Calculate the Mean Value.
b) Calculate the Average Deviation from the Mean
c) Calculate the: True Standard Deviation
d) Calculate the: Standard Deviation of the Mean.
e) If the correct value is 3.14159, calculate the: Percent Error
Answer:
a)[tex]MV=3.118[/tex]
b)[tex]AD=0.0416[/tex]
c)[tex]TSD=0.00302[/tex]
d)[tex]SD=0.00242[/tex]
e)[tex]%E=0.75%[/tex][tex]PE=0.75\%[/tex]
Explanation:
a)
The mean value equation is:
[tex]MV=\frac{\Sum x}{n}[/tex]
Where:
x represents the values of pin is the number of pi values[tex]MV=\frac{3.14+3.11+3.20+3.06+3.08}{5}[/tex]
[tex]MV=3.118[/tex]
b)
The average deviation equations is:
[tex]AD=\frac{\Sum |x-MV|}{n}[/tex]
[tex]AD=\frac{0.208}{5}[/tex]
[tex]AD=0.0416[/tex]
c)
The true Standard Deviation equation is:
[tex]TSD=\sqrt{\frac{\Sum (x-MV)^{2}}{n-1}}[/tex]
[tex]TSD=\sqrt{\frac{0.01208}{5-1}}[/tex]
[tex]TSD=0.00302[/tex]
d)
The Standard Deviation equation is:
[tex]TSD=\sqrt{\frac{\Sum (x-MV)^{2}}{n}}[/tex]
[tex]TSD=\sqrt{\frac{0.01208}{5}}[/tex]
[tex]SD=0.00242[/tex]
e)
The percent error equation is:
[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=\frac{|Measured value - True value|}{True value}[/tex]
[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=\frac{|3.118-3.14159|}{3.14159}*100[/tex]
[tex]%E=\frac{|Measured value - True value|}{|True value|}[/tex][tex]PE=0.75\%[/tex]
[tex]%E=0.75%[/tex]
I hope it helps you!
state what is meant by graviration potential at a point in an orbit 6.5×10^7
Explanation:
The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential,
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
Required:
At what rate is the magnetic field changing?
This question is incomplete, the complete question is;
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
At what rate is the magnetic field changing?
Answer:
the magnetic field changing at the rate of 9.33 m T/s
Explanation:
Given the data in the question;
Electric field E = 7 mV/m
radius r = 1.5 m
Now, from Faraday law of induction;
∫E.dl = d∅/dt
E∫dl = A( dB/dt )
E( 2πr ) = πr² ( dB/dt )
( 0.007 ) = (r/2) ( dB/dt )
( 0.007 ) = 0.75 ( dB/dt )
dB/dt = 0.007 / 0.75
dB/dt = 0.00933 T/s
dB/dt = ( 0.00933 × 1000) m T/s
dB/dt = 9.33 m T/s
Therefore, the magnetic field changing at the rate of 9.33 m T/s
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car? (Assume the bottom of the cliff is zero)
Answer:
Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.
modify other clauses and phrases in the sentence
refer to other sentences in the passage
make it hard to determine where the sentence ends
change the intended meaning of the sentence
Explanation:
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
the minimum charge on any object cannot be less than
Answer:
1.6 x 10^{-19} Coulombs
Explanation:
In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.
In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.
The minimum charge on any object such as an electron cannot be less than 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.
what is the main limitation of debye huckel theory
Answer:
Explanation:
For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. This is known as the Debye–Hückel limiting law.
THE ANSWER!!! Please
Answer:
I think -7 N. Netforce is 3N-10N= -7N
Explanation:
The only variable we found that affects the speed of a wave on a string was the tension of the string. How does this relate to how a musician tunes a stringed instrument?
Answer:
We know that in a string with two fixed points (like the one you will find in a guitar, where the string is "fixed" at the bridge and at the nut) the only thing that defines the speed (the frequency) at which the string vibrates is the tension (the length is also important, but for the 6 strings in the guitar all of them have the same length, but, as you know, when you press in a given fret the note changes, this happens because you are changing the length of the string, finally, the mass of the string is also important, but all the strings have almost the same mass, so we can ignore this).
Then for the open notes (the notes that you play when you don't fret any note) the only thing that defines the note that will sound is the tension of the string.
And a musician can tune the stringed instrument by changing the tension of each string using a tuner, which is the mechanism in the headstock of the instrument). As more tense is the string, higher will be the open note.
Rank the six combinations of electric charges on the basis of the electric force acting on q1.
a.
q1 = -1nC
q2= +1nC
q3= +1nC
b.
q1 = -1nC
q2= -1nC
q3= -1nC
c.
q1 = +1nC
q2= +1nC
q3= -1nC
d.
q1 = +1nC
q2= -1nC
q3= +1nC
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
A peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie plate moves through a distance of 208 in. Express the angular distance that the pie plate has moved through in revolutions, radians, and degrees.
Answer:
a) [tex]X_1=7.36rev[/tex]
b) [tex]X_2=46.22radians[/tex]
c) [tex]X_3=2649.6^o[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=9.00[/tex]
Distance [tex]x=208[/tex]
Generally the equation for circumference of a circle is mathematically given by
[tex]C=2 \pi r\\\\C=2*\pi*4.5[/tex]
[tex]C=28.3[/tex]
Therefore
Angular distance that the pie plate has moved through in revolutions is
[tex]X_1=\frac{x}{C}[/tex]
[tex]X_1=\frac{208}{28.3}[/tex]
[tex]X_1=7.36rev[/tex]
Generally Angular distance that the pie plate has moved through in radians is
[tex]X_2= 7.36rev* 2 \pi[/tex]
[tex]X_2=46.22radians[/tex]
Generally Angular distance that the pie plate has moved through in degrees is
[tex]X_3=7.36rev* 360[/tex]
[tex]X_3=2649.6^o[/tex]
The unit of distance used in astronomy is the light year, defined as the distance travelled by light in one calender year. How far away from earth (in km) is a star if its distance is quoted as 10 light years?
Answer:
9.7 trillion kilometers
Explanation:
A student must use an object attached to a string to graphically determine the gravitational field strength near Earth's surface. The student attaches the free end of the string to the ceiling and pulls the object-string system so that the string makes an angle of 5 degrees from the object's vertical hanging position. The student then releases the object from rest and uses a stopwatch to measure the time it takes for the object to make one complete oscillation. Which of the following is the next step that will allow the student to determine the gravitational field strength?
А) Repeat the experiment by adding additional mass to the object for multiple trials
B) Repeat the experiment by changing the length of the string for multiple trials
C) Repeat the experiment by changing the angle that the string makes with the object's vertical hanging position
D) Repeat the experiment by measuring the time it takes to make two oscillations, three oscillations, and additional oscillations for multiple trials
Answer:
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
Explanation:
The student is reacting a simple pendulum experiment where he can determine the value of the relationship of gravity with the expression
T = 2π [tex]\sqrt{\frac{L}{g} }[/tex]
let's analyze each statement
A) False. The mass is not a paramer of the period expression
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
C) False. The angle while it is small does not influence the period
D) True. By changing the number of oscillations the period does not change, so you can get the value of the pull of gravity.
We can see that the expressions B and d are true, the most exact value is obtained using procedure B since the graphs allow to reduce the errors
Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False
Answer:
False.
Explanation:
Suppose that we have an object that moves with constant acceleration A.
Then the acceleration of the object is defined by the equation:
a(t) = A
The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:
v(t) = A*t + V₀
Where V₀ is the velocity of the object at the time t = 0s.
Now, if we integrate it again, we will get the position equation:
p(t) = (1/2)*A*t^2 + V₀*t + P₀
Where P₀ is the initial position equation.
Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.
Part C
When only one color of light reflects from a piece of paper, what happens to the other colors of light?
Remember that light is energy, and energy cannot be created or destroyed.
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Answer:
the other colours get absorbed by the paper
Answer:
The other colors are absorbed by the paper and not reflected.
Which method of powering a vehiclewill help to reduce air pollution
using oil
using biofuels
using gasoline
using diesel fuel
Answer:
Using biofuels
Explanation:
The emissions of NOx and total VOCs lead to the formation of ozone in the troposphere, the main component of smog. ... Biofuels has a number of health and environmental benefits including improvement in air quality by reducing pollutant gas emissions relative to fossil fuels
(a) From an atomic point of view, why do you have to heat a solid to melt it? (b) If you have a solid and a liquid at room temperature, what conclusion can you draw about the relative strengths of their inter-atomic forces?
Answer:
A. & B. Heat energy is needed to convert solid into a liquid because heat energy increases the kinetic energy of the particles. The heat energy that it used to change 1 kg of solid into liquid at atmospheric pressure and at its melting point is called the latent heat of fusion.
Atoms of which two elements could combine with atoms of chromium (Cr) to
form ionic bonds?
O A. F.
B. Au
C. Se
D. Ti
E. Mg
Answer:
D and E
Ionic bonds are formed between metals and non-metals and both of those are metals
Define universal gravitational constant.
Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
A meter stick has a mass of 0.30 kg and balances at its center. When a small chain is suspended from one end, the balance point moves 28.0 cm toward the end with the chain. Determine the mass of the chain.
Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
A ball is thrown into the air with a velocity of 39 ft/s. Its height, in feet, after t seconds is given by s(t) = 39t − 16t2. Find the velocity (in ft/s) of the ball at time t = 1 second.
Answer:
7 ft/s
Explanation:
Applying,
V(t) = ds(t)/dt
Where V(t) = velocity of the ball at a given time
From the question,
Given: s(t) = 39t-16t²
Therefore,
V(t) = ds(t)/dt = 39-32t............. Equation 1
at t = 1 seconds,
Substitute the value of t into equation 1
V(t) = 39-32(1)
V(t) = 39-32
V(t) = 7 ft/s