Oxygen is obtained through various methods. Which of the following methods involves a chemical
change?
1. Electrolysis of water
2. Distillation of liquid air
3. Heating of KCIO,
02
1 and 2
1 and 3
Answer:
1
Explanation:
Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.
Heating a chemical will always cause a chemical reaction, which is why 3 is also correct
Some information as to why 2 is NOT correct.
2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.
as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)
Hope this helps.
Hope this helps.
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTA wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.
Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.
Answer: The correct statement is:
--> The voltage across each resistor is the same.
Explanation:
RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:
--> Resistors in parallel and
--> Resistors in series
RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.
How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination
Answer: hello tables and data related to your question is missing attached below are the missing data
answer:
a) I = I₁ = I₂ = I₃ = 0.484 mA
b) I₁ = 0.016 amps
I₂ = 0.0016 amps
I₃ = 7.27 * 10^-4 amps
c) I₁ = 1.43 * 10^-3 amp
I₂ = 0.65 * 10^-3 amps
Explanation:
A) magnitude of current for Part 1
Resistors are connected in series
Req = r1 + r2 + r3
= 3300 Ω ( value gotten from table 1 ) ,
V = 1.6 V ( value gotten from table )
hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA
The magnitude of current is the same in the circuit
Vi = I * Ri
B) magnitude of current for part 2
Resistors are connected in parallel
V = 1.6 volts
Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) + R3 ]
= [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]
= 87.30 Ω
For a parallel circuit the current flow through each resistor is different
hence the magnitude of the currents are
I₁ = V / R1 = 1.6 / 100 = 0.016 amps
I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps
I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps
C) magnitude of current for part 3
Resistors are connected in combination
V = 1.6 volts
Req = R1 + ( R2 * R3 / R2 + R3 )
= 766.66 Ω
Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps
magnitude of currents
I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps
I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps
why is the water drawn from the bottom of the dam rather than the top?
Answer:
because minerals can be gotten from the bottom
Explanation:
it's self explanatory
Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will
Solids cannot diffuse.
g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.
Answer:
have an increased resistance
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.
a. True
b. False
Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?
Answer:
[tex]X=0.0389m[/tex]
Explanation:
From the question we are told that:
Period of spring [tex]T_s=2.25s[/tex]
Initial Position of Mass [tex]x=0.0480m[/tex]
Final Mass period [tex]T_f=5.85s[/tex]
Generally the equation for the Mass location is mathematically given by
[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]
[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]
[tex]X=0.0389m[/tex]
Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
Diffuse reflection occurs when parallel light waves strike which surface? a mirror a rippling fountain a polished silver plate a still pond
Answer: a rippling fountain
Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.
Answer:
Answer
Correct option is
A
5×10
−6
tesla
I=5A
x=0.2m
Magnetic field at a distance 0.2 m away from the wire.
B=
2πx
μ
0
I
=
2π×0.2
4π×10
−7
×5
=10×5×10
−7
=5×10
−6
tesla
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
Two cylindrical resistors are made from copper. The first one is of length L and of radius r . The 2nd resistor is of length 6L and of radius 2r. The ratio of these two resistances R1/R2 is:
Answer:
[tex]R1/R2=\frac{2}{3}[/tex]
Explanation:
From the question we are told that:
1st's Length [tex]l=L[/tex]
1st's radius [tex]r=r[/tex]
2nd's Length [tex]l=6L[/tex]
2nd's radius [tex]r=2r[/tex]
Generally the equation for Resistance R is mathematically given by
[tex]R=\frac{\rho L}{\pi r^2}[/tex]
Therefore
[tex]R_1=\frac{\rho L}{\pi r^2}[/tex]
And
[tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]
Therefore
[tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]
[tex]R1/R2=\frac{2}{3}[/tex]
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
[tex]E_0=1.5033×10^{-10}\:\text{J}[/tex]
Explanation:
The rest energy [tex]E_0[/tex] of a proton of mass [tex]m_p[/tex] is given by
[tex]E_0 = m_pc^2[/tex]
[tex]\:\:\:\:\:\:\:=(1.6726×10^{-27}\:\text{kg})(2.9979×10^8\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:=1.5033×10^{-10}\:\text{J}[/tex]
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K
Answer:
c. 250k
Explanation:
The temperature of the sink is approximately 250 K.
To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:
Efficiency = 1 - (Temperature of Sink / Temperature of Source)
Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):
Efficiency = (Q_source - Q_sink) / Q_source
Efficiency = (2003 J - 100 J) / 2003 J
Efficiency = 1903 J / 2003 J
Efficiency = 0.9497
Now, plug the efficiency back into the first equation to solve for T_sink:
0.9497 = 1 - (T_sink / 500 K)
T_sink / 500 K = 1 - 0.9497
T_sink / 500 K = 0.0503
Now, isolate T_sink:
T_sink = 0.0503 * 500 K
T_sink = 25.15 K
Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.
To learn more about sink energy, here
https://brainly.com/question/10483137
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