please help!

[ISuckAtLife]

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K...I have some physics questions that I have been trying to tackle, but keep getting the wrong answer. Please help me

A 50 kg block resting on flat ground is attached by a hook at the center of its top face to a rope, which
is then pulled by a man such that it makes an angle of 30° with the horizontal. If the man pulls with a
force of 200 N and the coefficient of kinetic friction between the block and the ground is 0.2, what
horizontal acceleration does the block undergo?
A. 3.2 m/s2
B. 2.0 m/s2
C. 1.8 m/s2
D. 1.4 m/s2

For this one, I keep getting D 1.4, but apparently it is wrong. I drew this out and am pretty sure I am setting this up right so dunno why it's wrong.


A uniform meter stick held by one end is swung in an arc and released when the person’s arm is
horizontal, so that it moves initially away from the ground. About which point will it rotate as it flies
before striking the ground?
A. The end at which it was held
B. The 50 cm mark
C. The end furthest from the held pivot
D. It cannot rotate

For this one, I thought at first this might be a trick question and that D. It cannot rotate may be correct. Unfortunately, my instructor told me that's wrong and told me to look into why. My best guess is because it "initially moves away from the ground." Does this mean that B. The 50 cm mark is correct since that is the center of gravity?



An object of mass m (Object 1) moving with speed v collides head-on with a target object of mass 2m
(Object 2) initially at rest. If the collision is perfectly inelastic, what fraction of Object 1’s initial kinetic
energy is lost?
A. 1/9
B. 1/6
C. 1/3
D. 2/3

For this one, I am not entirely sure, but I got D. 2/3. Is this correct?


A point charge, Q = +1 mC, is fixed at the origin. How much work is required to move a charge, q =
+8 μC, from the point (0, 4 meters) to the point (3 meters, 0)?
A. 3.5 J
B. 6.0 J
C. 22.5 J
D. 40 J

For this, I keep getting C. 22.5 J. I drew this out, set it all up, was pretty confident about it, but apparently it is wrong. Not sure what I am missing here.


The following three step process refers to a simple RC circuit with a battery and an initially open
switch. Recall that the charge on a capacitor is given by Qc = CVc, and the energy stored in a charged
capacitor is given by PEc = ½CV
2.
Step 1: The switch is closed, allowing the capacitor to charge.
Step 2: After the capacitor has charged, a slab of dielectric material is inserted between the plates
of the capacitor and time passes.
Step 3: The switch is opened, and the dielectric is removed.
Which of the following describes the change in potential energy stored in the capacitor during each
step?
A. &#916;PE1 < 0; &#916;PE2 > 0; &#916;PE3 > 0
B. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 > 0
C. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 < 0
D. &#916;PE1 > 0; &#916;PE2 < 0; &#916;PE3 < 0

For this, I put C, which is wrong. Not sure why.


Optometrists use a linear scale of lens power, measured in diopters, to prescribe corrective lenses.
Lens power is given by the equation P = 1/f, where f is the lens focal length in meters. Sufferers of
myopia, or nearsightedness, have difficulty resolving distant objects because the lenses in their eyes
are too strong. If a myopic person has a prescription of –5 diopters, what image do her corrective
lenses create for distant objects?
A. an upright, real image about 20 cm in front of her eyes
B. an upright, virtual image about 20 cm in front of her eyes
C. an inverted, real image about 20 cm behind her eyes
D. an inverted, virtual image 5 m behind her eyes

For this, I was initially thinking C beceause of the sign convention of the formulas, but that was wrong. Perhaps B. since it's definitely 20 cm and because A upright, real is impossible.

Please help me anyone 😳

 

[Ssina]

K...I have some physics questions that I have been trying to tackle, but keep getting the wrong answer. Please help me

A 50 kg block resting on flat ground is attached by a hook at the center of its top face to a rope, which
is then pulled by a man such that it makes an angle of 30° with the horizontal. If the man pulls with a
force of 200 N and the coefficient of kinetic friction between the block and the ground is 0.2, what
horizontal acceleration does the block undergo?
A. 3.2 m/s2
B. 2.0 m/s2
C. 1.8 m/s2
D. 1.4 m/s2

For this one, I keep getting D 1.4, but apparently it is wrong. I drew this out and am pretty sure I am setting this up right so dunno why it's wrong.

sooo I think the answer would be C, since the force is being applied at an angle by the guy and has two components.. horizontal which opposes the friction and pulls it forward and the vertical, which opposes the force of gravity... so since it's opposing the gravity, the force of friction will be less, the guy is essentially pulling the box up a bit and forward (not enough obviously to lift it up the ground) by 200 sin 30=100... so the "new" normal force will be 500-100=400 and Ffriction= 0.2 * 400= 80N... then net force 173.2-80=50a.... a=1.8m/s2

A uniform meter stick held by one end is swung in an arc and released when the person's arm is
horizontal, so that it moves initially away from the ground. About which point will it rotate as it flies
before striking the ground?
A. The end at which it was held
B. The 50 cm mark
C. The end furthest from the held pivot
D. It cannot rotate

For this one, I thought at first this might be a trick question and that D. It cannot rotate may be correct. Unfortunately, my instructor told me that's wrong and told me to look into why. My best guess is because it "initially moves away from the ground." Does this mean that B. The 50 cm mark is correct since that is the center of gravity?

I think the answer is B, since just like when you throw a hammer, the center of mass (or in this ex 50cm) will follow a simple parabolic path, vs. everywhere else that will be in a loopy complicated trajectory. You are essentially applying torque, since it was held at the end, applied to the pivot point making the bar rotate.

An object of mass m (Object 1) moving with speed v collides head-on with a target object of mass 2m
(Object 2) initially at rest. If the collision is perfectly inelastic, what fraction of Object 1's initial kinetic
energy is lost?
A. 1/9
B. 1/6
C. 1/3
D. 2/3

For this one, I am not entirely sure, but I got D. 2/3. Is this correct?

I also got 2/3... since KEbefore=1/2mv2 and KEafter= 1/2 (3m) (v/3)2 and (KEbefore-KEafter)/KEbefore=2/3

A point charge, Q = +1 mC, is fixed at the origin. How much work is required to move a charge, q =
+8 &#956;C, from the point (0, 4 meters) to the point (3 meters, 0)?
A. 3.5 J
B. 6.0 J
C. 22.5 J
D. 40 J

For this, I keep getting C. 22.5 J. I drew this out, set it all up, was pretty confident about it, but apparently it is wrong. Not sure what I am missing here.

I got 6J for this one... so W=q(change in potential)... and potential lines run in circular rings with the given r around the point charge at (0,0)... so we are essentially bringing the second charge from 4m away to 3m away from the point change.. in this case the change in potential= k(q moving) (1/r2-1/r1)=9*10^9*8*10^-6(1/3-1/4)=6*10^3.... and then multiply that by 1*10^-3=6J

The following three step process refers to a simple RC circuit with a battery and an initially open
switch. Recall that the charge on a capacitor is given by Qc = CVc, and the energy stored in a charged
capacitor is given by PEc = ½CV
2.
Step 1: The switch is closed, allowing the capacitor to charge.
Step 2: After the capacitor has charged, a slab of dielectric material is inserted between the plates
of the capacitor and time passes.
Step 3: The switch is opened, and the dielectric is removed.
Which of the following describes the change in potential energy stored in the capacitor during each
step?
A. &#916;PE1 < 0; &#916;PE2 > 0; &#916;PE3 > 0
B. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 > 0
C. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 < 0
D. &#916;PE1 > 0; &#916;PE2 < 0; &#916;PE3 < 0

For this, I put C, which is wrong. Not sure why.

Also thought it would be C... at least i'm sure about 1 and 2, then I guess it's B?!

Optometrists use a linear scale of lens power, measured in diopters, to prescribe corrective lenses.
Lens power is given by the equation P = 1/f, where f is the lens focal length in meters. Sufferers of
myopia, or nearsightedness, have difficulty resolving distant objects because the lenses in their eyes
are too strong. If a myopic person has a prescription of &#8211;5 diopters, what image do her corrective
lenses create for distant objects?
A. an upright, real image about 20 cm in front of her eyes
B. an upright, virtual image about 20 cm in front of her eyes
C. an inverted, real image about 20 cm behind her eyes
D. an inverted, virtual image 5 m behind her eyes

For this, I was initially thinking C beceause of the sign convention of the formulas, but that was wrong. Perhaps B. since it's definitely 20 cm and because A upright, real is impossible.

Please help me anyone 😳

Myopia, so need a diverging lens, f=-20cm, and diverging lens will be SUV (small, upright, virtual).. so B sounds good to me too!
hopefully someone else will check this stuff as well!

 

[Ssina]

The following three step process refers to a simple RC circuit with a battery and an initially open
switch. Recall that the charge on a capacitor is given by Qc = CVc, and the energy stored in a charged
capacitor is given by PEc = ½CV
2.
Step 1: The switch is closed, allowing the capacitor to charge.
Step 2: After the capacitor has charged, a slab of dielectric material is inserted between the plates
of the capacitor and time passes.
Step 3: The switch is opened, and the dielectric is removed.
Which of the following describes the change in potential energy stored in the capacitor during each
step?
A. &#916;PE1 < 0; &#916;PE2 > 0; &#916;PE3 > 0
B. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 > 0
C. &#916;PE1 > 0; &#916;PE2 > 0; &#916;PE3 < 0
D. &#916;PE1 > 0; &#916;PE2 < 0; &#916;PE3 < 0

For this, I put C, which is wrong. Not sure why.

so it actually turns out that after you remove the dielectric the PE also increases:
since we removed the battery and the capacitor is now isolated in the circuite, there is no where for Q to go, so it will stay constant. The voltage, however, it will increase (V=Q/C, the C decreases by factor of K, so Voltage increase by factor of K)... when you remove the dielectric, the electric filed between the plates also increase, essentially increasing the potential on the plates as well....

so now we know that PE=1/2CV^2... the C decreased by K, and V increased by K (but here we have V^2) so PE increases by K also.... essentially you have to apply energy to pull the dielectric out

 

[ISuckAtLife]

so it actually turns out that after you remove the dielectric the PE also increases:
since we removed the battery and the capacitor is now isolated in the circuite, there is no where for Q to go, so it will stay constant. The voltage, however, it will increase (V=Q/C, the C decreases by factor of K, so Voltage increase by factor of K)... when you remove the dielectric, the electric filed between the plates also increase, essentially increasing the potential on the plates as well....

so now we know that PE=1/2CV^2... the C decreased by K, and V increased by K (but here we have V^2) so PE increases by K also.... essentially you have to apply energy to pull the dielectric out

Thank you so much for helping me. For this question, based on your reasoning it is B, right?

I have a few more that I have guesses on but was not sure on...

A sphere starts from rest atop a hill with a constant angle of inclination and is allowed to roll without
slipping down the hill. What force provides the torque that causes the sphere to rotate?
A. Static friction
B. Kinetic friction
C. The normal force of the hill on the sphere
D. Gravity

I think this is B. Kinetic friction. Is this right?


N resistors (N > 2) are connected in parallel with a battery of voltage Vo. If one of the resistors is
removed from the circuit, which of the following quantities will decrease?
I. The voltage across any of the remaining resistors
II. The current output by the battery
III. The total power dissipated in the circuit
A. I only
B. II only
C. II and III only
D. I, II, and III

Is this C? Since if you add a resistor in parallel, the current in the circuit increases and the power dissipated increases as well, right?


A 4 kg wooden block resting on an icy surface (so friction can be ignored) is attached to a horizontal
spring (L = 1 m, k = 100 N/kg), which is attached on the other end to a vertical wall. A 2 g bullet is fired
at a speed of 100 m/s into the wooden block, pushing it directly toward the wall. Which of the following
best approximates the period and amplitude of the resulting oscillations?
A. T = 1.2 s, A = 45 cm
B. T = 1.2 s, A = 1 cm
C. T = 0.83 s, A = 45 cm
D. T = 0.83 s, A = 21 cm

I got B. Is this right?


An electron (mass = m, charge = –e) is projected with speed v upward, in the plane of the page, into a
region containing a uniform magnetic field, B, that is directed into the plane of the page. Describe the
electron’s subsequent circular motion.
A. Clockwise rotation; radius of path = mv/(eB)
B. Counterclockwise rotation; radius of path = mv/(eB)
C. Clockwise rotation; radius of path = eB/(mv)
D. Counterclockwise rotation; radius of path = eB/(mv)

I got A for this, but am really unsure.

 

[MT Headed]

A sphere starts from rest atop a hill with a constant angle of inclination and is allowed to roll without
slipping down the hill. What force provides the torque that causes the sphere to rotate?
A. Static friction
B. Kinetic friction
C. The normal force of the hill on the sphere
D. Gravity

I think this is B. Kinetic friction. Is this right?

Kinetic friction requires two surfaces to slide past each other. It's not B.

 

[411309]

A sphere starts from rest atop a hill with a constant angle of inclination and is allowed to roll without
slipping down the hill. What force provides the torque that causes the sphere to rotate?
A. Static friction
B. Kinetic friction
C. The normal force of the hill on the sphere
D. Gravity

I think this is B. Kinetic friction. Is this right?

To elaborate, kinetic friction is what opposes a box being pushed across a surface. The same contact point touches the floor the entire time so it's kinetic friction. However for a ball, the ball is rolling not sliding. The ball is experiencing multiple points of contact with the ground and this is static friction. When a car drives along a road its static friction that opposes it, not kinetic.

Friction are forces that OPPOSE motion. So the static friction is acting to slow the sphere down, rather than cause it rotate. The normal force can be thought of the ground pushing up against the sphere. The sphere gets its acceleration from gravity. Acceleration causes the ball to rotate. So my educated guess is gravity.

 

[Ssina]

I think it's actually static friction... As MT Headed mentioned, no slipping then no kinetic friction, and the ball is rolling so there must be static friction.

Now gravity is acting on the center of mass, but static friction is at the surface between the ball and the inclined plane, so this is like a lever arm with the length of radius of the ball, to the center of mass (or the pivote point, point of rotation).... Providing torque. Essentially, if you overcome the static friction you will slide down instead of slipping!

 

[Dr. USMLE]

I think it's actually static friction... As MT Headed mentioned, no slipping then no kinetic friction, and the ball is rolling so there must be static friction.

Now gravity is acting on the center of mass, but static friction is at the surface between the ball and the inclined plane, so this is like a lever arm with the length of radius of the ball, to the center of mass (or the pivote point, point of rotation).... Providing torque. Essentially, if you overcome the static friction you will slide down instead of slipping!

I stumbled into this forum accidentally and thought I'd help...It is definitely not gravity. I agree that it is static friction.

 

[Dr. USMLE]

.

 

[MT Headed]

A sphere starts from rest atop a hill with a constant angle of inclination and is allowed to roll without
slipping down the hill. What force provides the torque that causes the sphere to rotate?
A. Static friction
B. Kinetic friction
C. The normal force of the hill on the sphere
D. Gravity

I think this question is deeper than the question writer wanted. Look at the question closely. It's asking which force provides the torque that causes the sphere to rotate. Torque is R x F x sin(theta). F is the force. R is the lever arm, i.e. the distance from the point of rotation to the point of the force. theta is the angle between F and R. A fair amount of this question boils down to choosing the proper point of rotation.

We could pick the center of the sphere as the point of rotation. Then the "R" of gravity would be zero. The "theta" of the normal force would be zero. And the product of "R", "F", and "theta" for the static force would be the only force that can produce a torque. Answer = A.

But why would we pick the center of the sphere as our point of rotation? A rolling sphere doesn't even roll about its center of mass. It rolls about its point of contact with the ground. Think about it - it's the only point of a rolling sphere that doesn't move; the rest of the sphere must rotate around this point or otherwise it would break up. If we choose the point of contact with the ground as the point of rotation, then the "R" of both the normal force and the static friction force is zero. The only force applying torque is gravity. Answer = D.

If I was diabolical I could choose a point where the normal force provides the necessary torque, and the other forces only provide translational motion. I'd be equally correct.

This question doesn't have an answer. If they want you to declare one force, they need to declare torque about what point in space. I'm sure the asker wanted you to answer Static Friction. I'd probably answer Gravity just out of spite, and then protest my case with the professor when it was marked wrong.

 

[Ssina]

It rolls about its point of contact with the ground. Think about it - it's the only point of a rolling sphere that doesn't move; the rest of the sphere must rotate around this point or otherwise it would break up.

I am not following how a sphere will rotate around the point of contact with the ground... the only point in the sphere that will follow a straight line down the rampe is the center of mass, while every other point rotates clockwise/counterclockwise (depending on the ramp) around the center while moving down.

 

[MT Headed]

I am not following how a sphere will rotate around the point of contact with the ground... the only point in the sphere that will follow a straight line down the rampe is the center of mass, while every other point rotates clockwise/counterclockwise (depending on the ramp) around the center while moving down.

Your translational vector "v" causes every point to move down the plane at velocity v. Your rotational velocity "omega" causes the point of contact to have a second velocity vector (v = omega/r) up the hill. The sum of these vectors is zero. It has to be zero, because otherwise the ball would be slipping. The center of the ball moves forward with a velocity v. The top of the ball would be moving forward with a velocity of 2v. The instantaneous center of rotation of a wheel is going to be the one point on the wheel that is not moving.

This wiki article probably explains it better than I can: http://en.wikipedia.org/wiki/Instan...nt_centre_of_a_wheel_rolling_without_slipping

You want to analyze rotational velocity about a point that is moving down the plane at a height "R" above the plane. That's fine. I'm just analyzing rotational motion about a different point on the ground that is also moving down the plane. It's also fine. Either point (the center of that ball, or the point of contact with the ground) makes analyzing some types of rotational questions easier.

Which force is responsible for the torque?
Which force is responsible for the change in angular velocity?
Well, that really depends on where you define your (moving) center of rotation to be. The only correct answer is E - not enough information given to solve this problem.

 

[TXKnight]

Maybe we are complicating it more than the question wants? I think the net force that's causing it to have a torque is gravity. Once again, maybe not...

 

[ISuckAtLife]

Maybe we are complicating it more than the question wants? I think the net force that's causing it to have a torque is gravity. Once again, maybe not...

Thanks again for helping me guys.

I asked my instructor for help on this and he was said it was not gravity, but never said what the answer was or how to approach it 😕

Also, was I right about the other ones Ssina? 😳

 

[Ssina]

Your translational vector "v" causes every point to move down the plane at velocity v. Your rotational velocity "omega" causes the point of contact to have a second velocity vector (v = omega/r) up the hill. The sum of these vectors is zero. It has to be zero, because otherwise the ball would be slipping. The center of the ball moves forward with a velocity v. The top of the ball would be moving forward with a velocity of 2v. The instantaneous center of rotation of a wheel is going to be the one point on the wheel that is not moving.

This wiki article probably explains it better than I can: http://en.wikipedia.org/wiki/Instan...nt_centre_of_a_wheel_rolling_without_slipping

You want to analyze rotational velocity about a point that is moving down the plane at a height "R" above the plane. That's fine. I'm just analyzing rotational motion about a different point on the ground that is also moving down the plane. It's also fine. Either point (the center of that ball, or the point of contact with the ground) makes analyzing some types of rotational questions easier.

Which force is responsible for the torque?
Which force is responsible for the change in angular velocity?
Well, that really depends on where you define your (moving) center of rotation to be. The only correct answer is E - not enough information given to solve this problem.

👍 didn't think of it this way....thanks for explaining it further!! In MCAT physics, we mostly assume everything has to do with the center (in terms of rotation, weight, etc), but I see the trick now... I'm sure they'll be more specific if it was going to be on the MCAT, but since this is a HW maybe OP should mention it to the professor

 

[ISuckAtLife]

👍 didn't think of it this way....thanks for explaining it further!! In MCAT physics, we mostly assume everything has to do with the center (in terms of rotation, weight, etc), but I see the trick now... I'm sure they'll be more specific if it was going to be on the MCAT, but since this is a HW maybe OP should mention it to the professor

Oh, when I said "instructor," I meant MCAT instructor. This is not HW...I finished my physics classes a while back, which may be part of the problem. 😉 These are just to help me prep for the MCAT.

 

[Ssina]

Oh, when I said "instructor," I meant MCAT instructor. This is not HW...I finished my physics classes a while back, which may be part of the problem. 😉 These are just to help me prep for the MCAT.

nevermind my bad! I did think they are all good questions for MCAT, no wonder why!
but what they said makes total sense... it's like a bar that you put on a ramp vertical to the surface... it will rotate down because of the gravity on the center mass since point of contact with the surface is not moving... but now if you hold the bar at the middle and drag it down now the friction is giving it torque at the point of contact... so they have to be more specific!

 

[Ssina]

N resistors (N > 2) are connected in parallel with a battery of voltage Vo. If one of the resistors is
removed from the circuit, which of the following quantities will decrease?
I. The voltage across any of the remaining resistors
II. The current output by the battery
III. The total power dissipated in the circuit
A. I only
B. II only
C. II and III only
D. I, II, and III

Is this C? Since if you add a resistor in parallel, the current in the circuit increases and the power dissipated increases as well, right?

that's what I got too... I think of it this way: by removing a R, since they are parallel, we increased the total R... since V doesn't change, because of batter, then I decreases and then Power decreases!

A 4 kg wooden block resting on an icy surface (so friction can be ignored) is attached to a horizontal
spring (L = 1 m, k = 100 N/kg), which is attached on the other end to a vertical wall. A 2 g bullet is fired
at a speed of 100 m/s into the wooden block, pushing it directly toward the wall. Which of the following
best approximates the period and amplitude of the resulting oscillations?
A. T = 1.2 s, A = 45 cm
B. T = 1.2 s, A = 1 cm
C. T = 0.83 s, A = 45 cm
D. T = 0.83 s, A = 21 cm

I got B. Is this right?

also got B

An electron (mass = m, charge = &#8211;e) is projected with speed v upward, in the plane of the page, into a
region containing a uniform magnetic field, B, that is directed into the plane of the page. Describe the
electron's subsequent circular motion.
A. Clockwise rotation; radius of path = mv/(eB)
B. Counterclockwise rotation; radius of path = mv/(eB)
C. Clockwise rotation; radius of path = eB/(mv)
D. Counterclockwise rotation; radius of path = eB/(mv)

I got A for this, but am really unsure.

got the same thing also, do the right hand rule but since it's electron, have to use the left hand... and since the electron will go in a circle, the force is essentially the centripetal force... mv^2/r=F(B)=evB so r=mv/eB

let me know if I messed up somewhere!

 

[ISuckAtLife]

Thanks Ssina...you are amazing!!

Are you a gen chem expert too?? I just had this one question that I stumbled on...

An indicator has Ka equal to 8 x 10–7. What is the ratio of the concentration of the acid form of the indicator
to the concentration of the basic form, [HIn] / [In–], in a buffer solution in which [H+] = 10–5 M?
A. 1/0.08
B. 8/1
C. 1/0.8
D. 1/8
E. 0.08/1

I ended up getting the answer 0.8/1, but that is not in the answer choices! So I picked E, which is wrong. I thought it might be C (like maybe if they wrote the question wrong since normally it is [A-]/[HA] instead of [HA]/[A-] as the question asks), but that is wrong too. 😕😕😕

 

[Ssina]

Thanks Ssina...you are amazing!!

Are you a gen chem expert too?? I just had this one question that I stumbled on...

An indicator has Ka equal to 8 x 10&#8211;7. What is the ratio of the concentration of the acid form of the indicator
to the concentration of the basic form, [HIn] / [In&#8211;], in a buffer solution in which [H+] = 10&#8211;5 M?
A. 1/0.08
B. 8/1
C. 1/0.8
D. 1/8
E. 0.08/1

I ended up getting the answer 0.8/1, but that is not in the answer choices! So I picked E, which is wrong. I thought it might be C (like maybe if they wrote the question wrong since normally it is [A-]/[HA] instead of [HA]/[A-] as the question asks), but that is wrong too. 😕😕😕

I actually suck at GChem, specially buffer stuff like this, but how about A:
the indicator: Ka= [H+][In-]/[HIn]=8x10-7, so given that [H+]=10-5, and that eventually give ratio of 1/0.08
I didn't so Henderson-Hasselbalch since you need the pKa of the acid that caused the pH to be, and we would get the ratio for that acid.... with that said, I'm not sure this is right or not! let me know if you figured it out

the pKa of the indicator is 6ish (~-log 10x10-7) , and the pH of solution is 5.... so pKa>pH so we have more protonated than deprotonated form of the indicator... so I would rule out D and E at least

 

[TXKnight]

I actually suck at GChem, specially buffer stuff like this, but how about A:
the indicator: Ka= [H+][In-]/[HIn]=8x10-7, so given that [H+]=10-5, and that eventually give ratio of 1/0.08
I didn't so Henderson-Hasselbalch since you need the pKa of the acid that caused the pH to be, and we would get the ratio for that acid.... with that said, I'm not sure this is right or not! let me know if you f

 

[TXKnight]

I actually suck at GChem, specially buffer stuff like this, but how about A:
the indicator: Ka= [H+][In-]/[HIn]=8x10-7, so given that [H+]=10-5, and that eventually give ratio of 1/0.08

I think Ssina is right
ph=pka-log (HA/A-)
pka is more like 7ish, thus we need log (HA/A-) be aprox 2 that gives the best choice 1/.08
....but maybe not

 

[kingte]

I think it's actually static friction... As MT Headed mentioned, no slipping then no kinetic friction, and the ball is rolling so there must be static friction.

Now gravity is acting on the center of mass, but static friction is at the surface between the ball and the inclined plane, so this is like a lever arm with the length of radius of the ball, to the center of mass (or the pivote point, point of rotation).... Providing torque. Essentially, if you overcome the static friction you will slide down instead of slipping!

Yes, you're right.

I agree with you.