Answers
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]E(r)_1 = [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]
Now when the outer diameter is doubled, the radius(b) is also doubled
So
[tex]E(r)_2 = [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]
Now
[tex]\frac{E(r)_2}{E(r)_1} = \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} = \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]
Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
Related Questions
2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she 4covers its 48.0 m length at a speed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0 m length at a speed of 5.00 m/s. How long does it take Suzette to make to class? Did Suzette beat the bell?
Answers
Answer:
62 secondsnoExplanation:
The total travel time Suzette experiences is the sum of the times in each hallway. Using
time = distance/speed
we can add the times.
(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)
= 10 s + 40 s + 12 s
= 62 s
It takes Suzette 62 seconds to get to class. She does not beat the bell.
N₂ + H₂
NH3
how do i balance this equation?
Answers
Answer:
N2 + 3H2 -----> 2NH3
Explanation:
Reactants side:
2 Nitrogen
5 Hydrogen
Products Side:
2 Nitrogen
5 Hydrogen
The first ionization potential for calcium (Z = 20, A = 40) is 6.11 eV. Singly-ionized calcium (Ca+) produces two very strong absorption lines in the Sun’s spectrum discovered by Joseph Fraunhofer in 1814, who named them "H" and "K" (he didn’t know they were from calcium, as this was >100 years before the development of quantum mechanics). Both lines always appear together, with lambda subscript H equals 3968 end subscript Å and lambda subscript K equals 3933 Å; hence they are called a "doublet
A. What is the speed of an electron that has just barely enough kinetic energy to collisionally ionize a neutral calcium atom? What is the speed of a calcium ion with this same kinetic energy?
B. What is the temperature T of a gas in which the average particle energy is just barely sufficient to ionize a neutral calcium atom?
C. The lower energy level of both lines is the ground state of Cat. What is the difference in energy in eV) between the two states that correspond to the upper energy levels of the Hand Klines, respectively? How does this compare to the energy of a calcium K photon? Can these two lines can be formed by transitions to upper energy levels with different principal quantum numbers (different n), or do they represent transitions with the same n but some different higher-order quantum number? Explain your reasoning based on your understanding of the general behavior of atomic energy levels (En).
Answers
Answer:
A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B) T = 4.7 10⁴ K, C) n₂ = 42
Explanation:
A) For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.
Let's reduce the units to the SI system
E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J
The kinetic energy of the electron is
K = ½ m v²
E₀ = K
v = √ 2E₀ / m
v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)
v = √ (2.14857 10¹²)
v = 1.47 10⁶ m / s
now the speed of a calcium ion is asked, let's find sum
m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg
v = √ (2E₀ / M)
v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)
v = √ (0.2994457 10⁸)
v = 0.5426 10⁴ m / s
B) the terminal energy of an ideal gas is
E = 3/2 kT
T = ⅔ E / k
T = ⅔ (9,776 10-19 / 1,381 10-23)
T = 4.7 10⁴ K
C) To calculate the energy of these lines we use the Planck expression
E = h f
where wavelength and frequency are related
c =λ f
f = c /λ
let's substitute
E = h c /λ
let's look for the energies
λ = 396.8 nm
E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹
E₁ = 5.0126 10⁻¹⁹ J
λ = 393.3 nm
E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹
E₂ = 5.0572 10⁻¹⁹ J
The difference in energy between these two states is
ΔE = E₂ -E₁
ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J
ΔE = 0.0446 10⁻¹⁹ J
let's reduce eV
ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
ΔE = 2.787 10⁻² eV
Now let's use Bohr's atomic model for atoms with one electron,
E = -13.606 Z² / n²
where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium
n = √ (13.606 Z² / E)
λ = 396.8 nm
E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV
n₁ = √ (13.606 20² / 3.132875)
n₁ = 41.7
since n must be an integer we take
n₁ = 42
λ = 393.3 nm
E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV
n₂ = √ (13.606 20² / 3.16075)
n₂ = 41.5
Again we take n as an integer
n₂ = 42
We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin
At an amusement park, a swimmer uses a water slide to enter the main pool. You may want to review (Pages 234 - 241) . Part A If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.81 m , what is her speed at the bottom of the slide
Answers
Answer:
Her speed at the bottom of the slide is 7.42 m/s
Explanation:
From the question,
The swimmer starts at rest, that is, her initial speed, u is 0 m/s.
Since she slides without friction and descends through a vertical height, then it is a free fall motion (due to gravity).
Also, from the question,
She descends through a vertical height of 2.81 m.
To determine her speed at the bottom of the slide, that is her final speed,
From one of the equations of motion for freely falling bodies
v² = u² + 2gh
Where v is the final speed
u is the initial speed
g is acceleration due to gravity (g = 9.8 m/s²)
and h is height
From the question,
u = 0 m/s
h = 2.81 m
Putting the values into the equation
v² = u² + 2gh
v² = 0² + 2×9.8×2.81
v² = 55.076
v =√55.076
v = 7.42 m/s
Hence, her speed at the bottom of the slide is 7.42 m/s.
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground
Answers
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration? (68.15 m/s)
Answers
v² - u² = 2 a ∆x
where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.
So
v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)
v² = 4645 m²/s²
v ≈ 68.15 m/s
An object at rest starts accelerating.
If it travels 20 meters to end up going 10 m/s.
what was its acceleration?
Variables:
Equation and Solve:
Answers
Answer:
We are given:
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
Solving for 'a'
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
The smokestack of a stationary toy tra in consists
of a vertical spring gun chat shoots a steel ball a meter
or so straight into the air-----so straight that the ball
always falls back into the smokescack. Suppose the
train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-
stack if shot from the moving train? What if the train
gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,
Answers
Answer:
i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Explanation:
The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
What is the answer to this ?
Answers
Engineers are using computer models to study train collisions to design safer
train cars. They start by modeling an elastic collision between two train cars
traveling toward each other. Car 1 is traveling north at 20 m/s and has a mass
of 12,745 kg. Car 2 is traveling south at 15 m/s and has a mass of 4,125 kg.
After the collision, car 1 has a final velocity of 3 m/s north. What is the final
velocity of car 2?
A. 56 m/s north
B. 56 m/s south
C. 38 m/s south
D. 38 m/s north
Answers
Answer:
did you get the answer???
Answer: 38 m/s north
Explanation:
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?
Answers
Answer:68.15m/s
Explanation:
Given:
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
Formula:
v₁²=v₁²+2a (x)
Set up:
=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]
Solution:68.15m/s
Please help me with this!
Answers
hope this helps good luck
Which would increase the speed of a sound wave?
O A wave passes from a solid to a liquid while remaining the same temperature.
The medium increases in temperature while remaining in the same phase.
The medium decreases in temperature while remaining in the same phase.
O A wave passes from a liquid to a gas while remaining the same temperature.
Answers
Answer:
The medium increases in temperature while remaining in the same phase
Explanation:
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
What is meant by the temperature coefficient of sound wave?The temperature coefficient of a sound wave is defined as the measure of increase in the velocity of the sound wave per unit rise in its the temperature.
Here,
The speed of a sound wave is affected by various factors in the medium through which it propagates. Among them, speed of the wave is mostly influenced by the temperature of the medium.
The speed of a sound wave in a medium is directly proportional to the square root of its absolute temperature. So,
v [tex]\alpha[/tex] √T
where v is thee speed of the sound wave and T is the absolute temperature.
Therefore,
As the temperature of the medium increases, the kinetic energy of the wave particles increases. Thus the speed of the sound wave is increased. As a result, the sound waves will move faster.
Hence,
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
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The plate is a model for how sunlight hits Earth’s surface. Which parts of Earth are most similar to the plate with an axis angle of 0°?
Answers
Answer: The parts closer to the equator.
Explanation:
The parts of the earth closer to the equator are similar to the plate, because they receive more sunlights than other parts. Which makes them hotter than any other regions in the earth. Example of such countries are Gabon, Uganda, Kenya, Maldives, the democratic republic of Congo, sao tome and principle e.t.c
Pressure and temperature ______ with depth below Earth’s surface.
Answers
Answer:
Pressure increases as you move deeper below earth's surface.
Tempurature increases as you move deeper below earth's surface.
Hope this helps!
Explanation:
Help me Please!!!!!!!
Answers
Do find the area of the triangle and that if the rectangle and add them together.
The area of the triangle is 25 and the rectangle is also 25 so the speed is 50 m/s
Hi, Please help.. I have assignments due tonight and I need someone to help me when a question I have generally..
Okay so if Density = Mass divided by Volume and I put that information into a calculator it comes out as
ex. 1.938773646 how do I make it look like something like this 1.4?
Answers
Answer:
did you tried putting it in standard form
Answer:
It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94
Explanation:
hope this helps!!
The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be
Answers
This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answers
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
The light bulbs are identical. Initially both bulbs are glowing. What happens when the switch is closed
Answers
Answer:
They turn off
Explanation:
What do mammoths and tigers need energy for
Answers
.............
Is the answer clockwise (CW) or counter clockwise (CCW) ?
Answers
Why is it better to use the metric system, rather than the English system, in scientific measurement?
A. The English system uses one unit for each category of measurement.
B. The metric system uses one unit for each category of measurement.
C. The English system uses consistent fractions that are multiples of 10.
D. The metric system utilizes a variety of number conversions.
Answers
A. The English system uses one unit for each category of measurement.
Answer:
A
Explanation:
An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the
Answers
Answer:
-z
Explanation:
The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it due to a magnetic field pointing in the +z direction.
As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.
Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons
Answers
it's unique for any element because it's determined by the atomic number and no two elements have the same atomic number
A crane uses a single cable to lower a steel girder into place. The girder moves with constant speed. The cable tension does work WT and gravity does work WG. Which statement is true
Answers
Explanation:
Work done by a force is given by :
[tex]W=Fd\cos\theta[/tex]
Where
F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.
In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.
We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.
I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.
The precision value of measuring tape is
1)0.1cm
2)0.1mm
3)1cm
4)0.01cm
Answers
C.1cm
Explanation:
precision is how close two or more measurements are to each other
4?
Explanation:
sorry im not sure but
You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)
The order goes like this:
rulers: 0.1cm or 1mm
measuring tape: 0.01cm or 0.1mm
vernier calipers: 0.001cm or 0.01mm
micrometer screw gauge: 0.0001cm or 0.001mm
((if i'm not wrong))
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answers
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
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Which of the following is the closest to the scientific fact
A. A hypothesis
B. A theory
C. An opinion
D. A prediction
Answers
A theory is the closest to a scientific fact (Option B).
A scientific theory is a well-sustained scientific idea that has been verified using the scientific method.A scientific theory can be refuted by the emergence of new lines of evidence against some aspect of this scientific statement.A hypothesis is a given explanation about a question that emerged by observing the natural world.In conclusion, a theory is the closest to a scientific fact (Option B).
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what phase changes take place when you are adding energy to the substance
Answers
Answer:
During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. Heat is added to a substance during melting and vaporization. Latent heat is released by a substance during condensation and freezing. Explanation: