Electrostatics and Circuits

1.3 C/s or A

Answer: A

Note on Current flow through series and parallel circuits

Passage A

Answer: B

1:√2

Passage A

The depth increases

Answer C

Passage B

Passage B

Loop rule

Passage B

0.1C/2s —→ 0.05 A

V=IR V= 0.05A (1000 ohms) = 50V

Answer= 50V

Increase in voltage because theres a decrease in resistance in series

Conductivity

indicates the ease with which electrons flow within the molecular strcuture of a material.

Conductivity is _______ proportional to resistivity

inversely

Suppose that a simple circuit ocmprising one voltage source and one metallic resistor yields current I. If the resistor were replaced with another resistor that is identical except that it has 75% lower conductivity, then I would……..

decrease in conductivity is increase in resistivity (increase R -resistance new) which means that R old is 25 percent of Rnew

Answer: D

increase in resistivity (due to decrease in conductivity)

V=IR

increase in Resistance increases the voltage

(note: it said current is conserved, so “I” can’t decrease as R increases (choice B)

Passage C

Passage C

Which change would result in the greatest decrease in the volumetric blood flow rate within a mesenteric vessel?

Answer: C

Passage C

What could be the units for “n” in equation 2?

Answer: C

Passage C:

A. Aorta

B. Arterial arcade

C. Venous arcade

D. Superior mesenteric artery

B. Arterial Arcade

Basically, the Q volumetric blood flow is conserved in series or parallels (like currents in sereis only) so the only thing that can change it is the P.

Increase in P increases R. Based on the image of graph,Arterial arcade has greatest blood pressure difference

.

Cofounding variable

uncontrolled variables that affect the dependent variable

B. Group with minimally invasive aortic catheterization to measure mesenteric blood rpessure

C. group fitted with noninvasive tail cuffs to measure systolic blood pressure

D. Group with laparotomy to measure renovascular blood pressure.

Passage C

Answer: B

in the original study, they’re measuring systolic blood pressure specifically in the mesenteric vessels (aorta, SMA, arterial arcade, venous arcade) using an invasive laparotomy.

So the dependent variable isn’t just “systolic blood pressure” in a general sense (e.g., brachial artery pressure via cuff). It’s mesenteric systolic blood pressure — a very specific measurement in a specific vascular bed.

That’s why:

• Choice C (tail cuffs) measures systemic systolic BP — different dependent variable → can’t isolate the effect of laparotomy on mesenteric pressures.

• Choice B (minimally invasive aortic catheter) measures mesenteric systolic BP without the laparotomy — same dependent variable, different level of the confounding variable → correct control.

So yes: same specific dependent variable, different procedure → that’s how you test if the procedure is a confounder.

Passage C

Answer: D

Passage C

Increase vascular flow resistance

Pressure drop across the msucle will increase

Blood flow trhough the gut will icnrease

Blood flow throughout the cardiovascular system will stop

Total vascular flow resistance will increase

Note: blood flow (Q) is conserved

Note: in capacitence series, resistance is added based on ex. ) 1/ (1/R1 +1/R2) while parallel you can directly add R1 + R2

1/C1 + 1/C2 (1/6 + 1/3) = 9/18 or ½—> 1/ (1/2) = 2 ——> C3 (9) + 2 = 11

Answer: 11 µF

Answer: 0C

original 2+2 = 4

Series 1/og or ¼ + ¼ ——> ½ —→ 1/(1/2) = 2 (new) so new: OG = 1:2

Passage D

Passage D

Passage D

X or Y?

Note: charge is directly proportional to force

so q =F= am a=v²/r

F=v²m/r =Q

decrease r increases F which increases Q which decreases m/q ratios and so its X

V= Ed

3000V = E (0.5m)

E= 6000V/m

or 6kN/C

V=Ed F=qE

F=qV/d

2000N/C * (2 1.6 ×10^-19) = 6.4 × 10^-16 N

Passage D

Answer: A

Why B is wrong

perpendicular to the ion’s velocity and parallel to the direction of the magnetic field

• “Perpendicular to velocity” is correct for magnetic force.

• “Parallel to magnetic field” is wrong — if force were parallel to B⃗B, then F⃗⋅B⃗≠0F⋅B=0, but from v⃗×B⃗v×B, the result is always perpendicular to B⃗B (so F⃗⋅B⃗=0F⋅B=0).

• This describes the electric force in some configurations, not magnetic.

---

Why C is wrong

parallel to both the ion’s velocity and the direction of the magnetic field

• Magnetic force is never parallel to velocity — if it were, it would do work and change speed, but magnetic force only changes direction, not speed (since F⃗⊥v⃗F⊥v).

• Being parallel to both v⃗v and B⃗B would require v⃗v and B⃗B to be parallel, but then v⃗×B⃗=0v×B=0, so no force.

• A force parallel to velocity means linear acceleration — particles would not curve in MS-MS chamber.

---

Why D is wrong

parallel to the ion’s velocity and perpendicular to the magnetic field

• Parallel to velocity → violates F⃗⊥v⃗F⊥v rule for magnetic fields.

• If a force were perpendicular to B⃗B but parallel to v⃗v, that would mean v⃗v is perpendicular to B⃗B (possible), but the force direction would still be perpendicular to both (v⃗×B⃗v×B), not parallel to v⃗v.

• Example: if v⃗v is along +x+x, B⃗B along +y+y, then F⃗F along +z+z (perpendicular to both, not parallel to v⃗v).

Passage E

Answer: Decreasing distance between the anode and cathode

Passage E

Answer: D

Passage F

Passage F

Answer A

Passage F

Passage F

Passage F

solve for F and direction

How is R shown on graph?

Note: circuit is parallel plate capacitor

Answer D

Power disipatted is greater in A or B? Reason?

Answer A

Passage G

Passage G

Passage G

Use table 1

110-70= 40

40= 10log (ratio)

4=logratio

10^4= ratio—→ 1000

Passage G

Answer A

Passage G

Answer C

Answer C

Loop rule: V- 1-3

Junction I1=I2+I3

V= IR

V=40

I=2

and solve for R total

then use R total to solve for R

R= 5

note: there are a few after that are not from electrostatic, accidentally put here

note

Answer C

visible spectrum

400 to 750nm

C is answer

isothermal thermodynamc process

adiabatic process

environment

isochoric process

isobaric process

answer c

use q=mcat with temp of nitrogen first to solve q then q=cat to find c with temp of sample

use tidal volume to solve

0.5L *1mol per 22.4L = 0.023 mol

particle B

Answer C