Jan 2003 Regents Physics Exam Explained James Wheeler

Jan 2003 Regents Physics Exam Explained James Wheeler

1) The worker exhibits a force with both magnitude and direction.

2) d = 90 m - 40 m = 50 m t = 20 s v=? v = d/t v = 50 m/20 s v = 2. 5 m/s

3) t = 3 s a = 9. 81 m/s 2 vi = 0 m/s d=? d = vit + (1/2)at 2 d = (0)(3 s) + (. 5)9. 81(3)2 d = 0 + (. 5)9. 81(9) d = 44. 145 m

4) Newton’s First law states that if any object is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

5) F 2 = 10 N a 1 = 5 m/s 2 m=? F = ma F/a= m m =(10 N)/(5 m/s 2) m = 2 g a 2 = 1 m/s 2 F 2= ? F 2=ma F 2 = (2 g)(1 m/s 2) F 2 = 2 N

6) g = Fg/m rise/run

7) m = 1200 kg v = 10 m/s t =. 1 s F=? F = ma F = m/(v/t) F= 1200 kg(10 m/s /. 1 s) F = 120, 000 N

8)

9) v = 12 m /s r = 30 m ac = v 2/r ac = (12 m/s)2/(30 m) ac = (144 m/s)/(30 m) ac = 4. 8 m/s 2

10) Centripetal acceleration is always towards the center. If it helps, remember that centripetal in Latin means “center seeking”.

11) μ = Ff/Fn Neither the force of friction or the normal force change which means that the coefficient of friction(μ) stays the same.

12) W = Fd Displacement(d) is a factor when it comes to work(W).

13) 3 m/s A p = mv p. A=p. B m Av A = m B v B (1. 2 kg)(v. A) = (1. 8 kg)(2 m/s) (1. 2 kg)(v. A) = 3. 6 v. A = 3 m/s B

14)

15) Δh =. 22 m mg = 15 N ΔPE= mgΔh ΔPE= (15 N)(. 22 m) ΔPE= 3. 3 J

16) KE = (1/2)mv 2 As an object is in freefall it accelerates, increasing its velocity and therefore increasing its kinetic energy.

17) q = 2. 5 x 10 -6 C W = 6. 3 x 10 -4 J V = W/q V = (6. 3 x 10 -4 J)/(2. 5 x 10 -6) V = 2. 5 x 102

18) F = 12 N v = (8 m)/(2 s) v = 4 m/s P = Fv P = (12 N)(4 m/s) P = 48 W

19) Opposite charges attract, like charges repel.

20) Fe= (kq 1 q 2)/r 2 As shown by the equation for electrostatic force, as the radius increases, the electrostatic force decreases exponentially.

21) Because the rod is positively charged, it wants to gain electrons, so it takes some from the neutral sphere.

22) Δq = 10 C T=5 s I = Δq/t I = (10 C)/(5 s) I=2 A

23) If the plates are parallel, then each point will experience the same magnetic field strength.

24) *Periodic waves transfer energy only.

25) P = VI As voltage(V) increases, power(P) increases as well.

26) v = 120 V I = 10 A t = 30 s W=? W = VIt W = (120 V)(10 A)(30 s) W = 36000 J W = 3. 6 x 104

27) Frequency(f): the # of cycles per second (Hz) 4/1 s = 4 Hz

28) The same point on each respective pulse A periodic wave is a series of pulses that are evenly timed.

30) f = 5. 09 x 1014 v = 2. 25 x 108 We must calculate the absolute index of refraction(n) n = c/v n = (3 x 108)/(2. 25 x 108) n = 1. 33

31) f = 5. 0 x 1014 c = 3 x 108 λ=? λ = v/f λ = (3 x 108)/(5. 0 x 1014) λ = 6 x 10 -7

32) The Doppler effect represents the relative change in frequency that is a result of a moving source of a wave.

33) A node is a point of zero displacement when two waves interact zero displacement

34) 1 e. V = 1. 6 x 10 -19 (1. 6 x 10 -19)(6. 0 x 106) 9. 6 x 10 -13

35) P = W/t P = J/s

Resultants:

a ay ax decreases ay increases θ ax

2. 2 Ib 2204. Ib

ET= PE + KE + Q ΔPE = mgΔh There is no change in potential energy because the track is purely horizontal, meaning there is no change in height(h). KE = (1/2)mv 2 There is no change in kinetic energy because the block is traveling at a constant velocity(v). There can be a change in internal energy(Q) only.

Utilize the electromagnetic spectrum in your reference table. X rays have the greatest frequency of all the options, and since there is a direct relationship between frequency and energy, X rays is the answer.

Since the energy in this system is purely potential, we can just calculate the total amount of work. d =. 5 m W = Fd F = 15 N W = (15 N)(. 5 m) W = 7. 5 J

Remember ohm’s law R=V/I (V) (I) Now that we know this graph measures resistance (R), and since a greater temperature means greater resistance, we know the answer should represent an increase in resistance.

Series Circuits I = I 1 = I 2 Req = R 1 + R 2 V = V 1+V 2 Req = 8 ohms + 8 ohms Req = 16 ohms V= 12 v I=? I= V/R I= 12 v/16 ohms I =. 75 A

The needle of a compass follows the magnetic field lines, which always travel from north to south.

The object is stationary v = d/t Constant velocity The object has a constant velocity, and then instantly decelerates. a = Δv/t Displacement grows exponentially as time increases.

Pes = (1/2)kx 2 2(Pes)/x 2 = k A B 2(Pes)/x 2 = k 2(. 4 N)/(. 04 m) 2(. 6 N)/(. 04 m) 20 = k 30 = k

Remember the law of conservation of charge 4 A - 1 A = 2 A 2 A – 1 A = 1 A 3 A – 2 A = 1 A

Check reference book for absolute indices. n 1 sin θ 1 = n 2 sin θ 2 (1)(sin 30) = (1. 5)(sin θ 2). 5 = 1. 5(sinθ 2). 333=sinθ 2

Medium Y, because according to the diagram it’s the densest medium, meaning that light travels the slowest in medium Y.

Superposition- the combined displacement of the two interfering waves is the algebraic sum.

v =19. 8 m/s θ = 30° /s m. 8 19 vy 30° vx sin 30° = vy/(19. 8 m/s). 5 =vy/(19. 8 m/s). 5(19. 8 m/s) = vy 9. 8 m/s = vy

a = -9. 81 m/s 2 a = Δv/t t = Δv/a t = (-9. 8 m/s – 9. 8 m/s)/ )/(-9. 81 m/s 2) t = (-19. 6 m/s)/(-9. 81 m/s 2) t=2 s

Voltmeters require a parallel connection V

The resistors are connected in parallel 1/Req = 1/4 Ω + 1/6 Ω 1/Req = 5/12 Ω Req = 2. 4 Ω

*Diffraction, is the spreading out of a wave into a region beyond an obstacle.

The strong force, or the strong nuclear force prevents the nucleus of a helium atom from flying apart.

L=1 m ρ(reference book) = 150. x 10 -8 A = 7. 85 x 10 -7 R = ρL/A R = (150 x 10 -8 Ωm)(1 m)/(7. 85 x 10 -7) R = 1. 91 Ω

R = 1. 91 Ω V = 1. 5 V I=? Ohm’s law I = (1. 5 V)/(1. 91 Ω) I =. 785 A

59 -61 62)

B, because the mass has the greatest velocity, where as the velocity at both point A and C would be 0.

A, because the height(h) is greatest.

Point C, because the change in length from the equilibrium position of the spring(x) is the greatest at point C.

6. 0 x 10 -15 N 8. 0 x 10 -15 N *always include arrowheads

6. 0 x 10 -15 N 8. 0 x 10 -15 N 10 cm

6. 0 x 10 -15 N 8. 0 x 10 -15 N 10 cm

6. 0 x 10 -15 N 8. 0 x 10 -15 N 10 cm θ 10 6 8

*check your reference table for coefficient of friction FN = 25 N =. 3 Ff = ?

Fn = 25 N Fg = 25 N

Fn = 25 N Ff = 7. 5 N Fa = 10 N Fg = 25 N

F = Fa – Ff F = 10 N – 7. 5 N F =2. 5 N Fn = 25 N Ff = 7. 5 N Fa = 10 N Fg = 25 N

Yes, the crate is accelerating because it is not in equilibrium and there is a net force acting on it. Fn = 25 N Ff = 7. 5 N Fa = 10 N Fg = 25 N

*Check your reference table Ei = -1. 51 e. V Ef = -3. 40 e. V Ephoton = Ei – Ef Ephoton = -1. 51 e. V – (-3. 40 e. V) Ephoton = 1. 89 e. V

Ephoton = 1. 89 e. V 1 e. V = 1. 6 x 10 -19 J Ephoton = (1. 89)(1. 6 x 10 -19) Ephoton = 3. 024 x 10 -19 J

E = 3. 024 x 10 -19 J f=?

or