Explanation:
F = ma
120.0 N = (3.00 kg) a
a = 40.0 m/s²
Which represents potential energy?
A) bread baking
B) lawnmower cutting grass
C) car racing downhill
D) weed whacker full of gas
If five 0.1ohm resistors are connected in parallel, the effective resistance is
Answer:
Effective resistance:
Rp = r/5 = 0.1/5 =0.02Ω
Explanation:
When the resistances are connected in parallel, the effective resistance becomes less than the smallest individual resistance. Thus smallest resistance is obtained when five0.1Ω resistors are connected in parallel.
which of the following is a disadvantage to asexual reproduction?
A. requires two different parents so not all the organisms Will be able to reproduce due to competition.
B. requires a lot of energy and therefore produces small numbers of organisms.
C. does not require much energy and therefore produces many offspring.
D. no genetic diversity mean organisms are vulnerable to disease and environmental changes.
What are some indicators that a chemical reaction is occurring? Check all that apply.
the formation of bubbles
a change in size
a change in color
the formation of a liquid
the formation of a precipitate
Answer:
the formation of bubbles
a change in color
the formation of a precipitate
Explanation:
Answer:
>the formation of bubbles
>a change in color
> the formation of a precipitate
How can you tell the difference between chemical sedimentary rocks and intrusive
igneous rocks?
Answer:
Sedimentary rocks are usually formed under water when grains of broken rocks are glued together while igneous rocks form when melted rock (magma or lava) cools and metamorphic are rocks that once were igneous or sedimentary rocks but have been changed by pressure and temperature.
Igneous rock is formed by the cooling of magma (molten rock) inside the Earth or on the surface. Sedimentary rocks are formed from the products of weathering by cementation or precipitation on the Earth's surface. Metamorphic rocks are formed by temperature and pressure changes inside the Earth.
Write a nuclear equation for the alpha decay of 231
91 Pa.
Answer:
231-91Pa>4-2a+227-89Ac
Explanation:
hope this helps
The equation for the alpha decay of [tex]^{231} _{91}Pa[/tex] is written as [tex]^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac[/tex].
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle.
The emission of this alpha particle reduces the mass number of the original atom by 4 and the atomic number by 2.
The alpha decay equation is written as;
[tex]^a_zX\ ---> \ ^4_2\alpha \ + \ ^{a-4} _{z-2}Y[/tex]
When the given atom ([tex]^{231} _{91}Pa[/tex]) under goes alpha decay, the resulting element is written using the following equation;
[tex]^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac[/tex]
Thus, the equation for the alpha decay of [tex]^{231} _{91}Pa[/tex] is written as [tex]^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac[/tex].
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how much thermal energy is absorbed by a copper water pipe with a mass of 2.3kg when its temperature is raised from 20.0 to 80.0?
Explanation:
q = mCΔt
q = (2.3 kg) (385 J/kg/K) (80.0°C − 20.0°C)
q ≈ 53,000 J
q ≈ 53 kJ
The thermal energy absorbed by 2.3 kg copper wire when its temperature raises from 20 to 80 degree Celsius is 53.13 J.
What is specific heat capacity?The specific heat capacity of substance is the heat energy required to raise its temperature by one degree Celsius per one gram. The specific heat capacity of metallic copper is 0.385 J/ °C g.
The calorimetric equation relating the heat energy q with the mass m, temperature difference ΔT and specific heat c is given by :
q = m c ΔT.
Given the mass of the copper wire = 2.3 Kg
temperature difference ΔT = 80 -20 = 60 °C.
specific heat of copper = 0.385 J/ °C g
q = 2.3 Kg × 0.385 J/ °C g × 60 °C
= 53.13 J
Therefore, the thermal energy required for the copper wire to raise its temperature is 53.13 J.
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Will there be noise cancellation in example A or
example B?
Answer:
b
Explanation:
Yes, there is a noise cancellation in example B.
What is noise cancellation?
Active noise control, also known as noise cancellation or active noise reduction, is a method of reducing unwanted sound by adding a second sound designed specifically to cancel the first.
Active noise cancellation employs more advanced technology to actively combat noise. Essentially, it detects and analyzes the sound pattern of incoming noise before producing a mirror "anti-noise" signal to cancel it out. As a result, you hear a significantly reduced level of noise.
As we understood that the unwanted noise is cancelled by the second sound wave which reduces or cancels the unwanted sound wave. Example B is showing that the other wave is cancelling the noise and making the sound more noise free.
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1. Explain how this picture is an example of kinetic energy.
2. What factors affect the energy of the ball?
3. What change could be made to give the ball more energy?
Answer:
1. He is putting force on the ball to throw it. 2. Him throwing the ball. 3. If he puts more force into throwing the ball it would give the ball more energy>>>
Explanation:
(1) The picture represents an example of kinetic energy because the ball is in motion.
(2) The factors affecting the energy of the ball are its mass and velocity.
(3) The change the could be made to give the ball more energy is to increase its mass or velocity.
What is kinetic energy?Kinetic energy is the energy of motion. In the given picture, the ball is in motion, so it represents an example of kinetic energy.
The factors that affect the energy of the ball include its mass and velocity. A ball with a larger mass or a higher velocity will have more kinetic energy.
To give the ball more energy, you could increase its mass or increase its velocity, which will in turn increase the kinetic energy of the ball.
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Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?
Answer: it's A hope this helps
Explanation:
Which observer would hear the louder sound?
a. The woman
C. They both do
b. The man
Answer:
B
Explanation:
the answer is b because the man has a deeper voice than the womans high pitched voice
Roshan makes the table below to describe how to draw a ray diagram for a convex lens.
What error did Roshan make?
The ray that goes through the center should bend and go through the focal point on the other side.
The ray that starts out parallel with the main axis should bend toward the axis and go through the focal point on the other side.
The ray that goes parallel to the main axis after passing through the lens should also be parallel from the object to the lens.
The rays in the table describe how rays are drawn for a concave lens rather than for a convex lens.
The error that Roshan made is: the rays in the table describe how rays are drawn for a concave lens rather than for a convex lens.
What is lens?A lens is an optical device that is used to focus or diverge light rays through refraction. It is made up of a transparent material, such as glass or plastic, with at least one curved surface.
Roshan's table contained an error because he described the ray diagram for a concave lens rather than a convex lens.
A concave lens's ray diagram differs from that of a convex lens because the concave lens diverges light while the convex lens converges light.
After passing through the lens, the ray should be parallel to the main axis from the object to the lens. This is due to the law of refraction, which states that the incident angle equals the refracted angle.
Thus, Roshan's error was describing the ray diagram for a concave lens rather than a convex lens.
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Roshan's error was in stating that a ray passing through the center of a convex lens should bend and go through the focal point on the other side, which is incorrect as such rays do not change direction.
Roshan's error in describing how to draw a ray diagram for a convex lens is in the statement that 'The ray that goes through the center should bend and go through the focal point on the other side.' In fact, rays that pass through the center of a convex (converging) lens continue in a straight line without bending. The correct rules are:
A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side.A ray passing through the center of either a converging or a diverging lens does not change direction.Therefore, the correct statements for a convex lens would be that a ray that starts out parallel with the main axis should bend toward the axis and go through the focal point on the other side, and that the ray passing through the center of the lens should not deviate from its path.
What is indicated by the slope of an acceleration vs. time graph?
Answer: The velocity of an object
Explanation: The slope of an acceleration vs. time graph is also the velocity of the object.
A cannon fires a cannonball. Explain how the cannonball gains kinetic energy.
A. The sound of the explosion propels the cannonball.
B. The thermal energy of the explosion propels the cannonball.
C. The chemical energy of the explosives transforms into kinetic energy.
D. The electrical potential energy of the explosives transforms into kinetic energy.
Answer:
C
Explanation:
I would try c but i'm not 100% positive
Answer:
c
Explanation:
its C because energy cant be destroy nor created its just can be transfer in a way such as the chemical solution that release energy to propel the cannonball thus the kinetic energy (movement)
1. The force between a pair of charges is 100 newtons. The distance between the charges is 0.01 meter. If one of the charges is 2 u 10-10 C, what is the strength of the other charge?
2. The force between two charges is 2 newtons. The distance between the charges is 2 u 10-4 m. If one of the charges is 3 u 10-6 C, what is the strength of the other charge?
Answer:
1. [tex]Q = 5.56 * 10^{-3} C[/tex]
2.[tex]Q = 2.97 * 10^{-12} C[/tex]
Explanation:
1. Electrostatic force is given as:
[tex]F = \frac{kqQ}{r^2}[/tex]
where k = Coulombs constant
q = charge of first charge
Q = charge of the second charge
r = distance between them
From the question:
F = 100 N
q = [tex]2 * 10^{-10} C[/tex]
r = 0.01 m
We need to find Q.
From the formula of force, we have that Q is:
[tex]Q = \frac{Fr^2}{kq}[/tex]
[tex]Q = \frac{100 * 0.01^2}{8.99 * 10^9 *2 * 10^{-10}} \\\\\\Q = \frac{0.01}{1.798}\\ \\\\Q = 0.00556 C = 5.56 * 10^{-3} C[/tex]
This is the charge, Q, of the second charge.
2. From the question:
F = 2 N
q = [tex]3 * 10^{-6} C[/tex]
r = [tex]2 *10^{-4} m[/tex]
We need to find Q.
Using the same formula for Q as in 1. above, we have that:
[tex]Q = \frac{2 * (2 *10^{-4})^2}{8.99 * 10^9 *3 * 10^{-6}} \\\\\\Q = \frac{8 * 10^{-8}}{26970}\\ \\\\Q = 2.97 * 10^{-12} C[/tex]
This is the charge, Q, of the second charge.
Answer:
(1) [tex]4.4835*10^{17}C[/tex]
(2) [tex]2.9748*10^{-12}C[/tex]
Explanation:
Force Between two charges is give by.
[tex]F=k\frac{Q_1 *Q_2}{r^2}[/tex] , here k is called coulomb constant and has value = [tex]8.967*10^9Nm^2/C[/tex].
(1) case, F =100N, r = 0.1m and [tex]Q_1[/tex]=[tex]2*10^{-10}C[/tex] substituting these values in above equation and solving for unknown gives us.
[tex]Q_2 = 4.4835*10^{17}C[/tex].
(2) Case, F = 2N, r = [tex]2*10^{-4}m[/tex] and [tex]Q_1=3*10^{-6}C[/tex].
again by substituting these in above equation and solving for unknown gives us.
[tex]Q_2=2.9748*10^{-12}C.[/tex]
a mountain climber weighs 42.0 N. If he climbs a hill 100m high. Calculate the work done in joules
Answer:
The answer is 4200 J.
Explanation:
The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :
W = F×D
F = 42N
D = 100m
W = 42 × 100
= 4200 J
The work done by the mountain climber who weighs 42.0 N and climbs a hill 100 meters high is 4200 joules, calculated using the formula for work done W = Fd.
The student's question relates to the concept of work done, which is a topic in Physics. To calculate work done, you can use the formula W = Fd, where W is work, F is force, and d is displacement (in the direction of force). In this case, the displacement would be the vertical height the mountain climber ascends, which is 100 meters.
The force is the weight of the mountain climber, which is given as 42.0 N. Applying the formula W = Fd, we get W = 42.0 N * 100 m, which equals 4200 joules. This represents the gravitational work done by the climber as he elevates himself against the pull of gravity over a vertical distance of 100 meters.
How would a small bar magnet be oriented when placed at position X
Answer:
D
Explanation:
Well opposites attract, don't they?
A bar magnet at position X will align according to external magnetic fields, typically aligning its north pole towards the geographic North Pole. When experimenting with two magnets, opposite poles attract while like poles repel, which can be visualized with iron filings arranging along a magnetic field.
When a small bar magnet is placed at position X, its orientation will be determined by the external magnetic fields present. If there is no significant external field other than the Earth's magnetic field, a freely suspended bar magnet would align with the Earth's magnetic field lines, where one pole (north-seeking) faces the geographic North Pole and the south-seeking pole faces the geographic South Pole.
However, if we follow instructions similar to those given—placing one bar magnet on the table and bringing the north end of another bar magnet close to the south end of the first, we expect to see attraction between the opposite poles. Conversely, if two like poles are brought close together, they will repel each other. These interactions are due to the magnetic field lines that emerge from the north pole of a magnet and enter at the south pole.
Bar Magnet and Magnetic Field Lines: Utilizing iron filings around bar magnets can visualize these lines. The filings align with the magnetic field, showing the typical dipole pattern. When two bar magnets are placed with the same poles facing each other, the filings will show a pattern indicating repulsion, with field lines diverging. In contrast, when opposite poles are near each other, filings will align to show attraction, with field lines converging.
Think about the motion of a roller coaster car, after completing a trip around the track
the car's displacement is ______
a. The length of the pathway the roller coaster car traveled.
b. The starting point minus the finishing point
с. Zero
d. Half the distance because of vector addition
Answer:
it should be B let me know if i am wrong
Explanation:
Which kind of body is the largest?
1. asteroids
2. comets
3. meteoroids
4. all are the same size
Answer: ( A ) - asteriods
Explanation: because most asteriods are in the asteriods belt , and the largest asteroid in Ceres , which is 583 miles across , still holding the record .
Asteroids refer to the type of body which is the largest in this scenario and is denoted as option 1.
What is an Asteroid?These are rocky objects which orbit the sun and are smaller which is why they aren't referred to as planets.
They are however bigger than the meteoroids and comets which makes it the most appropriate choice.
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Humans start from a single cell. Eventually, as cells divide, they start to differentiate, or specialize. Some cells are specialized for movement of the human body. What type cells would these be?
Answer:
interphase im sure
Explanation:
(PLEASE HELP) In a solution such as salt water, the component that does the dissolving is called the
A. solute
B. solvent
C. suspension
D. colloid
Answer:
B) Solvent
Explanation:
This is because the solute is dissolved in the solvent.
In the example of salt water, the water is the solvent and the salt is the solute.
The solute is substance being dissolved.
The solvent is what the solute is being dissolved in.
In a wave is generated in a slinky and the wave is transmitted to a string that is less dense then the Slinky the Transmitted wave will be
Answer: Your answer is "erect".
Explanation:
Got this one right on the quiz.
What is the voltage drop across the 20.0 resistor?
A. 1.50 V
B. 20.0 V
C. 30.0 V
D. 90.0 V
Answer: B i did a test
The voltage drop across the 20.0-ohm resistor is 30 V. Correct Option is 3.
When resistors are connected in series, the total resistance (R_total) is the sum of individual resistances. In this case, the total resistance is:
R_total = 60 ohm + 20 ohm = 80 ohm
According to Ohm's law, the voltage drop across a resistor is proportional to its resistance. Therefore, the voltage drop across the 20.0-ohm resistor can be calculated using the ratio of its resistance to the total resistance, multiplied by the total voltage.
Voltage drop across the 20.0-ohm resistor:
V_drop = (Resistance of the 20.0-ohm resistor / Total resistance) * Total voltage
V_drop = (20 ohm / 80 ohm) * 120 V
V_drop = (1/4) * 120 V
V_drop = 30 V
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In which situations is heat flow present? Check all that apply.
A substance feels warm.
One substance does not have a temperature.
Two substances have the same temperature.
One substance transfers some of its thermal energy to another substance.
Two substances differ in temperature.
Answer:
A. A substance feels warm.
D. One substance transfers some of its thermal energy to another substance.
E. Two substances differ in temperature.
Explanation:
Which is an example of radiation?
A ray of light traveling in air strikes a flat 2.00 cm thick block of glass (n = 1.50) at an angle of 19.0 ◦ with the normal.?
a) Trace the light ray through the glass, and find the angle of refraction for light passing from air to glass. Answer in units of ◦.
b) Find the angle of refraction for light passing from glass to air. Answer in units of ◦.
Answer:
(a). [tex]\theta_2 = 12.5^o[/tex]
(b). [tex]\theta_1 = 19.0^o[/tex]
Explanation:
(a).
Snell's law says
[tex]n_1sin(\theta_1) = n_2sin(\theta_2)[/tex]
which in our case gives
[tex](1.00)sin(19.0^o) = (1.50)sin(\theta_2)[/tex]
Solving for [tex]\theta_2[/tex] gives
[tex]\dfrac{(1.00)sin(19.0^o)}{1.50} = sin(\theta_2)[/tex]
[tex]\theta_2 = sin^{-1}[\dfrac{(1.00)sin(19.0^o)}{1.50}][/tex]
[tex]\boxed{\theta_2 = 12.5^o}[/tex]
which is the angle of refraction from air to glass.
(b).
When the light ray goes from glass to back to air again, the angle of refraction equals the angle with which it had entered the glass because of the symmetry in Snell's law:
[tex](1.00)sin(\theta_1) = (1.50)sin(12.5^o)[/tex]
[tex]\boxed{\theta_1 = 19.0^o}[/tex]
The result is that the outgoing ray (one that refracted out from glass to air) and the tracing of the incoming ray (one that refracted into the glass) are parallel to each other.
Which of the following are steps of the inquiry process?
The steps of the inquiry process in historical inquiry are: developing questions, analyzing sources, developing evidence, and constructing claims.
Explanation:The steps of the inquiry process in historical inquiry are as follows:
Develop compelling and supporting questions that guide the inquiry process.Locate and analyze historical primary and secondary sources.Analyze the historical sources to develop historical evidence.Construct claims about the past, also known as historical interpretations, based on the evidence.These steps allow historians to engage in the process of historical inquiry and gain a comprehensive understanding of the past.
By what factor will the electric force between two charged objects change if the charge on both objects is quadrupled
A. 1/4
B. 4
C. 8
D. 16
What type of wave is this
Answer:Longitudinal wave/compression wave
Explanation:?
A block is initially at rest on top of an inclined ramp that makes an angle θ0 with the horizontal. The base of the ramp has a length of D . After the block is released from rest, it slides down the ramp onto a rough horizontal surface until it comes to rest at a position x=4D from the base of the ramp, as shown in the figure. There is negligible friction between the block and the inclined ramp, while the coefficient of kinetic friction between the block and the rough horizontal surface is μb .
(a) On the axes below, sketch and label graphs of the following quantities as a function of the position of the block between x=−D and x=4D . Calculations for values for the vertical axis are not necessary, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The gravitational potential energy Ug of the block-Earth system
(b) The block is released from the top of a new ramp that has a base length of 2D, but still makes an angle θ0 with the horizontal. A student is asked to predict whether the final horizontal position of the block will be twice as far from the base of the ramp compared to when it was released from the original ramp. The student reasons that since the block will be released from a new height that is twice as high as the original height, the block will have more energy when it reaches the base of the ramp, so it will slide farther along the right surface before stopping at a position x=8D.
i. Which aspects of the student’s reasoning, if any, are correct? If no aspect of the student’s reasoning is correct, write “none”.
ii. Which aspects of the student’s reasoning, if any, are incorrect? If no aspect of the student’s reasoning is incorrect, write “none”.
(c) Derive an equation for the new final position of the block. Express your answer in terms of D.
(d) In the following question, refer to the relationships written in part (c), not just the final answer obtained by manipulating those relationships.
For any correct aspects of the student’s reasoning identified in part (b)(i), how is the student’s reasoning expressed by your mathematical relationships in part (c) ?
For any incorrect aspects of the student’s reasoning identified in part (b)(ii), how do your relationships in part (c) correct the student’s incorrect reasoning?
a) i) See graph in attachment
ii) See graph in attachment.
b) i) All of them
ii) none
c) [tex]s=\frac{4D tan \theta}{\mu_s}[/tex]
d) See explanation below
Explanation:
a)
Find in attachment the graph showing the kinetic energy and the potential energy versus the position, x.
i)
Between x = -D and x = 0, the block is sliding down along the ramp. The kinetic energy of the block at any time is given by:
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m is the mass of the block
v is its speed
At the beginning, the block's kinetic energy is zero, because the speed is initially zero: since v = 0, KE = 0.
As the block slides down, the kinetic energy increases, because the speed of the block increases; at x = 0 (end of the ramp), all the initial energy has been converted into kinetic energy, which is now maximum.
Then, the block slides along the flat, rough surface; as friction does (negative) work on the block, the speed of the block decreases, and so also its kinetic energy decreases, becoming zero when x = +4D (when the block comes to a stop).
ii)
The potential energy of the block is given by:
[tex]GPE=mgh[/tex]
where
m is the mass of the block
g is the acceleration due to gravity
h is the height of the block above the ground
At the beginning (x = -D) the potential energy is maximum since the block is at maximum height.
When the block slides down (between -D and 0), the height h decreases, therefore the potential energy decreases as well, until reaching 0 when x = 0 (end of the ramp).
After x = 0, the block slides along the rough surface; however, its potential energy here no longer changes, as the height h dors not change (the surface is horizontal).
b)
Here, the block is released from the top of a new ramp, which has a base length of 2D (instead of D) but same angle as before: therefore, the initial height of the ramp is twice that in part a). This also means that the initial (potential) energy of the block in this case is twice the GPE of part a):
[tex]GPE'=2GPE[/tex]
As a result, when the block reaches the end of the ramp at x = 0, it will have twice the kinetic energy it had before:
[tex]KE'=2KE[/tex]
The stopping distance of an object moving with accelerated motion is proportional to its initial kinetic energy:
[tex]s\propto KE[/tex]
Therefore, this means that here the stopping distance of the block will be twice that of part a (which was 4D), so the block will stop at x = +8D.
So, all aspects of the student's reasoning are correct.
c)
Let's call [tex]E[/tex] the initial total energy of the block at the top of the ramp.
In situation a), the initial total energy is
[tex]E=GPE=mgh = mgD tan \theta[/tex]
where [tex]h=Dtan \theta[/tex] is the height of the ramp.
And so the kinetic energy at the bottom of the ramp is
[tex]KE=E[/tex]
We can rewrite the kinetic energy so that
[tex]\frac{1}{2}mv^2=E \rightarrow v \sqrt{\frac{2E}{m}}\\\rightarrow v=\sqrt{2gD tan \theta}[/tex]
For an accelerated motion, the stopping distance can be found using the equation
[tex]v'^2-v^2=2as[/tex]
where
[tex]v'=0[/tex] is the final speed of the block
[tex]a=-\mu_b g[/tex] is the acceleration due to friction
So we find
[tex]s=\frac{-v^2}{2a}=\frac{(2gD tan \theta)}{\mu_s g}=\frac{2D tan \theta}{\mu_s}[/tex] (1)
In situation b), the initial height of the block is
[tex]h=2D tan \theta[/tex]
So the final stopping distance becomes (1)
[tex]s=\frac{4D tan \theta}{\mu_s}[/tex]
d)
We can see that the formula derived in part c) depends only on:
- The initial height of the ramp, which is [tex]Dtan \theta[/tex] in part a) and [tex]2D tan \theta[/tex] in part b)
- The coefficient of friction in the rough part, [tex]\mu_s[/tex]
- The angle of the ramp, which remains the same in the two cases
Therefore, all the correct aspects identified by the student in his reasoning are found in the fact that the final stopping distance is proportional to the initial energy of the block, which is proportional to initial height of the block, and since this is twice in part b) compared to part a), therefore the stopping distance is also twice in part b).
Refer the below solution for better understanding.
Given :
Block has initial velocity, u = 0.
Base of the ramp has a length of D.
Negligible friction between the block and the inclined ramp.
Coefficient of kinetic friction between the block and the rough horizontal surface is [tex]\rm \mu_ b[/tex].
Solution :
a) Graph of the following quantities is attached below.
i) We know that the Kinetic energy is,
[tex]\rm KE = \dfrac{1}{2}mv^2[/tex]
at x = 0, KE is zero because initial velocity is zero but kinetic energy increases as velocity increases. And at an instant kinetic energy becomes maximum because velocity is maximum.
When the block slides along the flat, rough surface, friction acts on the block, the speed of the block decreases, so kinetic energy is also decreases and becomes zero at x = 4D.
ii) We know that potential energy is given by,
PE = mgh
At x = -D , height of the block is maximum therefore potential is also maximum at x = -D. The potential energy decreases as well, until reaching 0 when x = 0.
b) The initial height of the ramp is twice that in part a). This also means that the initial (potential) energy of the block in this case is twice the PE of part a). As a result, when the block reaches the end of the ramp at x = 0, it will have twice the kinetic energy it had before.
Stopping distance is proportional to its initial kinetic energy:
Therefore, the stopping distance of the block will be twice that of part a), so the block will stop at x = 8D.
Therefore, all aspects of the student's reasoning are correct.
c) Initial total energy is,
E = PE
[tex]\rm E= mgh = mgDtan\theta[/tex]
Kinetic energy at bottom of the ramp is,
KE = E
[tex]\rm \dfrac{1}{2}mv^2 = mgDtan\theta[/tex]
[tex]\rm v = \sqrt{2gDtan\theta}[/tex]
We know that,
[tex]\rm v'^2 = u^2 +2as[/tex]
here,
[tex]\rm v' = 0[/tex] and [tex]\rm a = -\mu_bg[/tex] (acceleration due to friction). So,
[tex]\rm s = \dfrac{-u^2}{2a}[/tex]
[tex]\rm s = \dfrac{2gDtan\theta}{\mu_bg}[/tex]
[tex]\rm s = \dfrac{2Dtan\theta}{\mu_b}[/tex] ----- (1)
In section b) h = [tex]\rm2Dtan\theta[/tex]
Now equation (1) becomes,
[tex]\rm s = \dfrac{4Dtan\theta}{\mu_b}[/tex]
d) We can see that the formula derived in part c) depends only on:
The initial height of the ramp, which is [tex]\rm Dtan\theta[/tex] in part a) and [tex]\rm 2Dtan\theta[/tex] in part b).
The coefficient of friction in the rough part, [tex]\mu _s[/tex].
The angle of the ramp, which remains the same in the two cases.
Therefore, all the correct aspects identified by the student in his reasoning are found in the fact that the final stopping distance is proportional to the initial energy of the block, which is proportional to initial height of the block, and since this is twice in part b) compared to part a), therefore the stopping distance is also twice in part b).
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