Answer:
Water Vapour, Carbon dioxide, methane,nitrous oxide,ozone.
David and Fiona want to know about how mass affects the motion of an object. David uses three objects with masses of 2 kg, 3 kg, and 8 kg, writes out his procedure, and records his observations on the force needed to move the objects. Fiona is investigating three different objects using observation. What is the difference in the way they approached this question?
A. Only Fiona gathered evidence from her investigation.
B. David used scientific articles for evidence.
C. David's observations can be replicated by repeating his procedure.
D. Fiona's observations can be replicated by observing different objects.
Answer: the answer is D if its not i'm really sorry
If the total translational kinetic energy of the molecules of oxygen in a container is 15 J at room temperature, what is the total rotational kinetic energy of these molecules
Answer:
zero(0)
Explanation:
if a molecule is moving along a straight path, there is zero rotational speed and that component give zero moment in the direction of the molecule's driving force
The ratio of total rotational kinetic energy to the total translational kinetic energy is 3 : 2. Thus, the rotational kinetic energy of oxygen with translational kinetic energy of 15 J is 10 J.
What is rotational kinetic energy?Rotational kinetic energy is the energy generated by virtue of the rotational motion of a molecule. The degree of freedom of a compound or molecules is the sum of its translational degree of freedom, rotational degree of freedom and vibrational degree of freedom.
For a diatomic molecule, there are three translational degrees of freedom along x, y and z directions. Similarly there will be two rotational degree of freedom along the molecular axis.
The ratio of rotational degree of freedom to that of translational motion is 3/2.
Thus Te / Re = 3/2
15 J/ Re = 3/2
Rotational kinetic energy Re = 30/3 =10 J.
Therefore, the rotational kinetic energy oxygen would be 10 J.
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Consider 5.00 mol of liquid water. (a) What volume is occupied by this amount of water? The molar mass of water is 18.0 g/mol. (b) Imagine the molecules to be, on average, uniformly spaced, with each molecule at the center of a sma11 cube. What is the length of an edge of each sma11 cube if adjacent cubes touch but don't overlap? (c) How does this distance compare with the diameter of a molecule?
This question involves the concepts of density, volume, and mass.
(a) The volume occupied by this amount of water is "90 m³".
(b) The length of an edge of each small cube is "83 μm".
(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.
(a)
The volume can be found using the following formula:
[tex]V = \frac{nM}{\rho}[/tex]
where,
V = volume occoupied = ?
n = no. of moles = 5 mol
M = molar mass = 18 g/mol
[tex]\rho[/tex] = density of water = 1 g/m³
Therefore,
[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]
V = 90 m³
(b)
First, we will find the volume of an individual molecule:
[tex]V_i =\frac{V}{nN_A}[/tex]
where,
[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol
Therefore,
[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]
Vi = 3 x 10⁻²³ m³
This volume can be given as a volume of the sphere:
[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]
r = 4.15 x 10⁻⁷ m
Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)
d = 8.3 x 10⁻⁷ m = 83 μm
since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.
Edge Length = L = d
L = 8.3 x 10⁻⁷ m = 83 μm
(c)
The edge length is equal to the diameter of the molecule.
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If you have a light bulb connected to a circuit and you increase the current, what happens to the light bulb and the power (assuming the voltage remains the same)? A. The light bulb shines brighter and the power decreases. B. The light bulb shines brighter and the power increases. C. The light bulb gets dimmer and the power decreases. D. The light bulb gets dimmer and the power increases.
Answer:
B. The light bulb shines brighter and the power increases.
Explanation:
Power is the product of voltage and current . Increasing the current at a constant voltage will increase the power.
Increasing the power means more heat and therefore more light must be emitted.
Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stretch by 4 cm, what is the ratio of the potential energy of S1 and S2? Use g = 10 m/s^2
Select one:
a. 1:2
b. 4:1
c. 1:4
d. 1:3
e. 2:1
Answer:
potential energy = mgh
= 400÷1000 × 10× 4÷100
= 0.4 × 10 × 0.04
=4/10 ×10×4/100
= 4/10 × 4/10
=16/100
= 0.16 joules
m1 (400) stretches 4cm
m1 (100g) stretches 1cm
so, m2(800g) stretches 8 cm
potential energy of m2 = mgh
= 800/1000 ×10×8/100
= 0.8 × 0.8
=8/10 ×8/10
= 64/100
=0.64 joules
Ratio of s1 to s2
16/100 ÷ 64/100
= 1:4 ( answer)
what is a capacitor and capacitance.
Answer:
Capacitor is a device which used to store energy .
Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.
Explanation:
There are 270 students and teachers going on a field trip to a science center. If each school bus holds 54 people, how many buses are needed?
groundwater returning to earths surface comes up through a ___?
Answer:
Groundwater Returning to earths Surface come Up Though A drought
Explanation:
THIS DOES NOT NEED TO BE EXPLAINED IT'S EZZZ
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship
The length of the ship in terms of Emily's equal steps is 84.
The given parameters;
equal steps forward = 210equal steps backward = 42Defined parameters:
Let the constant velocity of the ship = VLet the velocity of Emily = VsLet the length of the ship = dLet the time of motion, = tThe velocity of Emily when moving forward in the direction of the ship:
[tex]V_s_1 = \frac{210}{t}[/tex]
The velocity of Emily when moving in opposite direction to the ship:
[tex]V_s_2 = \frac{42}{t}[/tex]
The constant velocity of the ship:
[tex]V = \frac{d}{t}[/tex]
Apply relative velocity formula to determine the length of the ship:
For forward (same direction) motion:
[tex](V_s_1 - V)t = d[/tex]
For backward (opposite direction) motion:
[tex](V_s_2 + V)t = d[/tex]
[tex](V_s_1 - V)t = (V_s_2 + V)t\\\\V_s_1 - V = V_s_2 + V\\\\\frac{210}{t} - \frac{d}{t} = \frac{42}{t} + \frac{d}{t} \\\\\frac{210}{t} - \frac{42}{t} = \frac{d}{t} + \frac{d}{t} \\\\\frac{210 - 42}{t} = \frac{d+ d}{t} \\\\\frac{168}{t} = \frac{2d}{t} \\\\2d = 168\\\\d = \frac{168}{2} \\\\d = 84[/tex]
Thus, we can conclude that the length of the ship in terms of Emily's equal steps is 84.
"Your question is not complete, it seems to be missing the following information;"
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship
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You are traveling along a freeway at 65 mi/h. You suddenly skid to a stop because of congestion in traffic. Where is the energy that your car once had as kinetic energy before you stopped
The work and energy theorem allows finding the result for where the kinetic energy of the car is before stopping is:
The energy becomes:
An important part in work on discs. A part in non-conservative work due to friction.
Work is defined by the scalar product of force and displacement.
W = F . d
Where the bold indicate vectors, W is work, F is force and d is displacement.
The work energy theorem relates work and kinetic energy.
W = ΔK = [tex]K_f - K_o[/tex]
In this case the vehicle stops therefore its final kinetic energy is zero, consequently the work is:
W = - K₀
Therefore, the initial kinetic energy that the car has is converted into work in its brakes. In reality, if assuming that there is friction, an important part is transformed into non-conservative work of the friction force, this work can be seen in a significant increase in the temperature of the discs on which the work is carried out.
In conclusion, using the work-energy theorem we can find the result for where the kinetic energy of the car is before stopping is:
The energy becomes:
An important part in work on the discs. A part in non-conservative work due to friction.
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The fundamental cause of a variable star's change in luminosity is that the star's _____ is changing.
A Vector that starts from
Origin is called what?
the momentum of a car before the crash is 22500kg m/s. the car stops in 0.14s. what is the average force on the car during the crash?
Answer:
what is the full questio n because I can't see the momentum
Balance the following chemical equation:
CO2 + VH20 - CH1206+ voz
[
~
CO₂+
Answer:
Explanation:
Start with Carbon and assume we only get 1 sugar molecule from the process.
you have 6 carbons in the sugar on the right, so you need 6 carbons on the left which only come from CO₂
6 CO₂
you have 12 hydrogen atoms in the sugar on the right, so you need 12 hydrogen atoms on the left which only come from H₂O. At 2 hydrogen atoms per water molecule means you need 6 waters.
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆
you are supplied with 12 oxygen from the CO₂ and 6 oxygen from the H₂O, but you only need 6 oxygen for the sugar. That means there are 12 oxygen remaining which will become 6 O₂ molecules
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
6. A 15.53 kg bag of soil falls 5.50 m at a construction site. If all the energy is retained by the soil in the bag, how much will its temperature increase
The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.
The given parameters:
mass of the bag, m = 15.53 kgheight of fall, h = 5.5 mspecific heat capacity of soil, C = 0.200 kcal/kg ⁰CApply the principle of conservation of energy as follows;
[tex]mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}[/tex]
Convert the value of C in kcal/kg ⁰C to J/kg ⁰C
1 kcal = 4180 J
[tex]0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C[/tex]
The increase in the temperature of the soil is calculated as follows;
[tex]\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C[/tex]
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The flow of ground water is
Answer:
In hydrogeology, groundwater flow is defined as the "part of streamflow that has infiltrated the ground, entered the phreatic zone, and has been (or is at a particular time) discharged into a stream channel or springs; and seepage water." It is governed by the groundwater flow equation.
49. A particle starts from rest at time t=0 and movies along the x axis. if the net force on is proportional to t its kinetic energy is proportional to?
Answer:
F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]
Explanation:
In pic
_________________
(hopet his helps can I pls have brainlist (crown)☺️)
A phone cord is 2.28 m long. The cord has a mass of 0.2 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.849 s. What is the tension in the cord?
The characteristics of the speed of the traveling waves allows to find the result for the tension in the string is:
T = 10 N
The speed of a wave on a string is given by the relationship.
v =[tex]\sqrt{\frac{T}{\mu } }[/tex]
Where v es the velocty, t is the tension ang μ is the lineal density.
They indicate that the length of the string is L = 2.28 m and the pulse makes 4 trips in a time of t = 0.849 s, since the speed of the pulse in the string is constant, we can use the uniform motion ratio, where the distance traveled e 4 L
v = [tex]\frac{d}{t}[/tex]
v = [tex]\frac{4 L}{t}[/tex]
v = [tex]\frac{4 \ 2.28 }{0.849}[/tex]
v = 10.7 m / s
Let's find the linear density of the string, which is the length of the mass divided by its mass.
μ = [tex]\frac{m}{L}[/tex]
[tex]\mu = \frac{0.2}{2.28}[/tex]
μ = 8.77 10⁻² kg / m
The tension is:
T = v² μ
Let's calculate
T = 10.7² 8.77 10⁻²
T = 1 0 N
In conclusion using the characteristics of the velocity of the traveling waves we can find the result for the tension in the string is:
T = 10 N
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While forming a 1.5kg aluminum statue, a metal smith heats the aluminum to 2700 degrees C, pours it into a mould, and then cools it to a room temperature of 23.0 degrees C. Calculate the thermal energy released by the aluminum during the process.
Answer is supposed to be: 4.7*10^6 j/kg?
The heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
The given parameters:
Mass of the statues, m = 1.5 kgFinal temperature of the status, t₂ = 2700 CTemperature when it is in the mould, t₁ = 23 ⁰CSpecific heat capacity of aluminum, C = 900The heat released by the aluminum during the process is calculated as follows;
[tex]Q = mc \Delta t \\\\Q = 1.5\times 900 \times (2700 - 23)\\\\Q = 3.6 \times 10^6 \ J[/tex]
Thus, the heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
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I need to choose a theme for my physics assignment My experiment is finding g
How to find g (acceleration due to gravity)
Solution:We know,
Acceleration due to gravity (g)
[tex] = \frac{GM}{ {R}^{2} } [/tex]
where, G = Gravitational constant
[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]
M = Mass of the earth
[tex] = 6 \times {10}^{24} \: kg[/tex]
R = Radius of the earth
[tex] = 6.4 \times {10}^{6} m[/tex]
Putting these values of G, M and R in the above formula, we get
[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]
So, the value of acceleration due to gravity is
[tex]9.8m/s ^{2} [/tex]
Hope it helps.
Do comment if you have any query.
Astone has a mass of 200 grams. When it is immersed in a measuring cylinder of water,the water rises 100 ml.What is the density of the stone
Answer:
2 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]d = \frac{m}{v} \\ [/tex]
m is the mass
v is the volume
From the question
m = 200g
v = 100 mL
We have
[tex]d = \frac{200}{100} = 2 \\ [/tex]
We have the final answer as
2 g/mLHope this helps you
The less dense areas created as a sound wave propagates are called
A. rarefactions✅
B. troughs
C. crests
D. compressions
Answer:
C. crests
don't trust me though
Answer:
The less dense areas created as a sound wave propagates are called rarefactions.
Explanation:
Hoped this helped.
3. What is Newton's 1st Law?
A For every action there is an equal and opposite reaction
B. Acceleration depends on two variables, the mass of the object and the amount of
force
C. An object at rest will stay at rest, an object in motion will stay in motion, unless an
unbalanced force acts upon it.
O D. The amount of matter in an object
Answer:
A
Explanation:
for every action there is an equal and opposite reaction
HIHIHHHHHHHHIHIHIIHIHHHHIHIHIHIHIHIHIHIHIHIHIHIHIHIIH
Answer:
Ok thanks for the points though
Explanation:
A boat is headed north at a velocity of 8 km h-1. A strong wind is blowing whose pressure on the boat’s superstructure causes it to move sideways to the west at a velocity of 2 km h-1. There is also wave present that flows in a direction 30° south of east at a velocity of 5 km h-1. What is the velocity of the boat?
The velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.
The given parameters;
velocity of boat northwards, = 8 km/hvelocity of the boat westwards, = 2 km/hvelocity of the wind, = 5 km/h 30° south of eastThe resultant vertical velocity of the boat is calculated as;
[tex]v_y = 8 \ - \ 5 sin(30)\\\\v_y = 8 - 2.5\\\\v_y = 5.5 \ km/h[/tex]
The resultant horizontal velocity of the boat is calculated as;
[tex]v_x = -2 \ + 5cos(30)\\\\v_x = 2.33 \ km/h[/tex]
The resultant boat velocity is calculated as follows;
[tex]v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.33^2 + 5.5^2} \\\\v = 5.97 \ km/h[/tex]
The direction of the velocity;
[tex]\theta = tan^{-1} (\frac{v_y}{v_x} )\\\\\theta = tan^{-1}(\frac{5.5}{2.33} )\\\\\theta = 67 \ ^0[/tex]
Thus, the velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.
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Which is a force that wears away landforms? Select three options.
A. weathering
B. erosion
C. humans
D. clouds
E. light
A 0.015 kg marble sliding to the right at 0.225 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 0.180 m/s. After the collision, the first marble moves to the left at 0.180 m/s. Find the velocity of the second marble after the collision.
Explanation:
[tex]if \: the \: masses \: of \: the \:two \: marbles\: equal \\ then \: each \: velocity \: sharing \: with \: other \: velocity[/tex]
¿Por qué el origen de ciertos metales se relaciona con la dinámica terrestre?
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
An archer's bow is drawn at its midpoint until the tension in the string is 0.842 times the force exerted by the archer. What is the angle between the two halves of the string
Consult the attached free body diagram.
If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces
• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0
• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0
since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).
We only really need the first equation. Simplifying it, we get
F [archer] - T cos(θ) - T cos(θ) = 0
F [archer] - 2T cos(θ) = 0
F [archer] = 2T cos(θ)
cos(θ) = F [archer] / (2T)
We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say
T = 0.842 F [archer]
and from this we have
cos(θ) = F [archer] / (2 • 0.842 F [archer])
cos(θ) = 1/1.684
cos(θ) ≈ 0.593
Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.