Year End Review Double-click Word Art Put Subject of Game Here and put your topic here or click once and delete.

Kinematics

Newton

Energy

Electricity

Magnetism

5 pt. 5 pt. 5 pt. 5 pt. 5 pt. 10 pt. 10 pt. 10 pt. 10 pt. 10 pt. Kinematics Newton Energy Electricity Magnetism 5 pt. 5 pt. 5 pt. 5 pt. 5 pt. 10 pt. 10 pt. 10 pt. 10 pt. 10 pt. 15 pt. 15 pt. 15 pt. 15 pt. 15 pt. 20 pt. 20 pt. 20 pt. 20 pt. 20 pt. 25 pt. 25 pt. 25 pt. 25 pt. 25 pt.

1-25 The graph below is a velocity vs time graph. What is the distance covered in the first 10 seconds? 0 10 20 30 40 50 60 Velocity (m/s) 0 2 4 6 8 10 12 14 16 18 20 Time (s) 5 pt.

Calculate the area under the graph for the first 10 seconds. 0 10 20 30 40 50 60 Velocity (m/s) 0 2 4 6 8 10 12 14 16 18 20 Time (s) . Calculate the area under the graph for the first 10 seconds. Area = 310 m/s*s or 310 m 5 pt.

What is the acceleration of a car that goes from 0 to 100 m/sec in 2 minutes? (answer in m/sec2) 10 pt.

2-20A What is the acceleration of a car that goes from 0 to 100 m/sec in 2 minutes? (answer in m/sec2) a = v / t and, t = 2 min (60sec/min) = 120 sec so… a = (100m/s - 0m/s)/120 sec a = 0.833 m/sec2 10 pt.

1-15 A dragster accelerates down the track at a constant rate. It starts from rest and crosses the finish line moving at a speed of 250 mph. What is its average velocity? 15 pt.

vavg = (250mph + 0mph)/2 vavg = 125 mph 15 pt. A dragster accelerates down the track at a constant rate. It starts from rest and crosses the finish line moving at a speed of 250 mph. What is its average velocity? vavg = (250mph + 0mph)/2 vavg = 125 mph 15 pt.

3-25 1-20 How far would a sprinter run if they start at rest and end the race moving at a velocity of 20 m/sec? Assume they run for 10 seconds. 20 pt.

vavg = (20m/sec + 0m/sec)/2 = 10 m/sec d = vavgt = (10m/sec)(10sec) How far would a sprinter run if they start at rest and end the race moving at a velocity of 20 m/sec? Assume they run for 10 seconds. vavg = (20m/sec + 0m/sec)/2 = 10 m/sec d = vavgt = (10m/sec)(10sec) d = 100 meters 20 pt.

3-20 How much time would it take to run 100 meters if you start at 10m/s and accelerate at 2 m/sec2? 25 pt.

3-20A How much time would it take to run 100 meters if you start at 10m/s and accelerate at 2 m/sec2? d = vot + 1/2at2 100m = 10m/s*t + 0.5*2m/s2*t2 Use quadratic formula t = 6.18 seconds 25 pt.

You and a friend push a heavy 2-20 You and a friend push a heavy crate across the floor at a constant speed. If you push with a combined force of 100 N, how hard is friction pushing back on the crate? A) less than 100 N B) 100 N exactly C) more than 100 N D) not enough info 5 pt.

You and a friend push a heavy crate across the floor at a constant speed. If you push with a combined force of 100 N, how hard is friction pushing back on the crate? A) less than 100 N B) 100 N exactly C) more than 100 N D) not enough info 5 pt.

3-10A What is the weight of a 50-kg object as it falls at a speed of 100 m/sec in a vacuum? 10 pt.

The speed of the object has no effect on its weight, so… What is the weight of a 50-kg object as it falls at a speed of 100 m/sec in a vacuum? The speed of the object has no effect on its weight, so… W = mg = (50 kg)(9.8 m/sec2) W = 490 Newtons 10 pt.

3-20A A 2000-kg car accelerates down the road. The engine drives it with a force of 5000 N, and friction from the road is 1400 N. What is the car’s acceleration? 15 pt.

3-20A A 2000-kg car accelerates down the road. The engine drives it with a force of 5000 N, and friction from the road is 1400 N. What is the car’s acceleration? Fnet = ma and Fnet = 5000 N – 1400 N So… Fnet = 3600 N = (2000 kg)a a = 3600 N/2000 kg = 1.8 m/sec2 15 pt.

3-25A A 75-kg parachutist is falling through the air – while experiencing an air drag of 200 N. If they fall from rest for 10 seconds in this manner, how fast are they moving? 20 pt.

3-25A A 75-kg parachutist is falling through the air – while experiencing an air drag of 200 N. If they fall from rest for 10 seconds in this manner, how fast are they moving? v = a t so we must find a… Fnet = Weight – Drag = ma So… (75 kg)(9.8 m/s2) – 200 N = (75 kg)a 535 N/75 kg = 7.13 m/s2 = a v = (7.13 m/s2)(10 sec) = 71.3 m/sec 20 pt.

2-25 A rocket of mass 250-kg is pushed by a thrust force of 10,000-N through 12 kilometers. How fast is it moving when the engine shuts off? 25 pt.

2-25A A rocket of mass 250-kg is pushed by a thrust force of 10,000-N through 12 kilometers. How fast is it moving when the engine shuts off? F = ma 10,000N = 250kg * a a = 40m/s2 vf2 = vi2 + 2ad vf2 = 02 + 2*40m/s2*12,000m vf = 980 m/sec 25 pt.

1-15A 3-5 How much work is done by an engine when it pushes a 2000kg car with a force of 1450 N through a distance of 200 meters? 5 pt.

1-15A How much work is done by an engine when it pushes a 2000kg car with a force of 1450 N through a distance of 200 meters? Work = Force x distance Work = (1450 N)(200 m) = 290,000 J 5 pt.

4-20A 3-10 When jumping up and down on a trampoline, the 450-N jumper leaves the trampoline with a KE of 3800 J. How high do they travel? 10 pt.

KEinitial = GPEfinal = mgh 3800 J = (450 N) h When jumping up and down on a trampoline, the 450-N jumper leaves the trampoline with a KE of 3800 J. How high do they travel? KEinitial = GPEfinal = mgh 3800 J = (450 N) h h = (3800 J)/(450 N) = 8.44 m 10 pt.

4-25A 3-15 A 2-kg ball slides from rest down a 5-m tall incline as shown. When it is halfway down, how fast is it going? 15 pt.

(2kg)(10m/s2)(5m) = 1/2 (2kg)v2 + (2kg)(10m/s2)(2.5m) 3-15A 4-25A A 2-kg ball slides from rest down a 5-m tall incline as shown. When it is halfway down, how fast is it going? GPEinitial = KE + GPE (2kg)(10m/s2)(5m) = 1/2 (2kg)v2 + (2kg)(10m/s2)(2.5m) 100 J = v2 + 50 J v = √(50) = 7.1 m/sec 15 pt.

3-20 A spring is compressed 0.4m by a 24kg mass. The mass is removed and the spring is stretched 0.8m. How much energy is in the spring? 20 pt.

3-20A A spring is compressed 0.4m by a 24kg mass. The mass is removed and the spring is stretched 0.8m. How much energy is in the spring? F = kx 24kg * 10m/s2 = k * 0.4m k = 600 N/m PES = ½ kx2 PES = 192 J 20 pt.

3-25 How much power does the engine of a 2300-kg elevator put out if it climbs through 38-m in one minute? 25 pt.

3-25A How much power does the engine of a 2300-kg elevator put out if it climbs through 38-m in one minute? GPE = (2300 kg)(10 m/s2)(38 m) = 874,000 J Power = GPE / Time = (874,000 J)/(60 sec) Power = 14,567 Watts 25 pt.

#1 F = 6 blugs + + #2 F = ? blugs + + 5 pt. 4-5A 2-15 4-5 If two equal charges (as shown in #1 below) have a force of 6 blugs between them, what will be the force between the charges in #2? #1 + + F = 6 blugs + #2 + F = ? blugs + + 5 pt.

#1 F = 6 blugs + + #2 F = 24 blugs + + 5 pt. 4-5A 2-15 If two equal charges (as shown in #1 below) have a force of 6 blugs between them, what will be the force between the charges in #2? #1 + + F = 6 blugs + #2 + F = 24 blugs + + 5 pt.

Two protons are located 0. 07m from each other Two protons are located 0.07m from each other. What is the force of electrostatic repulsion between them? Proton charge = +1.6x10-19 C 10 pt.

10 pt. F = k q1q2 / r2 F = (9x109 N2m2/C2)(1.6x10-19C)2 (0.07m)2 Two protons are located 0.07m from each other. What is the force of electrostatic repulsion between them? Proton charge = +1.6x10-19 C F = k q1q2 / r2 F = (9x109 N2m2/C2)(1.6x10-19C)2 (0.07m)2 F = 4.70x10-26 Newtons 10 pt.

+ What is the electric field at point P, 2.5m from the 3μC charge? P 4-15 4-15 What is the electric field at point P, 2.5m from the 3μC charge? P + 15 pt.

+ What is the electric field at point P, 2.5m from the 3μC charge? P E = kQ / r2 E = 9E9*3E-6 / (2.5)2 E = 4320 N/C to the RIGHT 15 pt.

What is the voltage across the 60Ω resistor in the circuit below? 20 pt.

What is the voltage across the 60Ω resistor in the circuit below? Resistance of parallel section is 20Ω There are 2 equal voltage drops So each is 10v 20 pt.

4-25 A 600 Ω and 300 Ω are in parallel. If they are hooked up to a 20 volt battery, how much power is generated by the battery? 25 pt.

A 600 Ω and 300 Ω are in parallel. If they are hooked up to a 20 volt battery, how much power is generated by the battery? 1/Rt = 1/R1 + 1/R2 + 1/R3 Rt = 200Ω Pt = Vt2 / Rt Pt = 20v2 / 200Ω Pt = 2 Watts 25 pt.

If an Electron travels into this magnetic field as shown, what is the direction of the force on the charge? - 5 pt.

If an Electron travels into this magnetic field as shown, what is the direction of the force on the charge? - Force 5 pt.

A charge of 3mC, 5mg travels at 1000m/s through a magnetic field of 20 Tesla. What is the radius of the resulting arc? X X X X X X X X X X X X X + 10 pt.

A charge of 3mC, 5mg travels at 1000m/s through a magnetic field of 20 Tesla. What is the radius of the resulting arc? F = QvB sin q = mv2 / r QB = mv / r r = mv / QB r = 5*10-9kg*1000m/s / (3mC*20T) r = .083m 10 pt.

A 20cm wire has a current of 10amps A 20cm wire has a current of 10amps. It passes through a 15cm wide magnetic field of 20 Tesla. What is the force on the wire? 5-15 I 15 pt.

A 20cm wire has a current of 10amps A 20cm wire has a current of 10amps. It passes through a 15cm wide magnetic field of 20 Tesla. What is the force on the wire? F = ILB sinq F = 10amps*.15m*20T F = 30 N (into the board) 15 pt.

2 parallel wires each carry 3 amps of current in the same direction 2 parallel wires each carry 3 amps of current in the same direction. If they are 1m long and 2cm apart, how strong is the force between them? 5-20 20 pt.

2 parallel wires each carry 3 amps of current in the same direction 2 parallel wires each carry 3 amps of current in the same direction. If they are 1m long and 2cm apart, how strong is the force between them? B = moI/2pr B = 4p*10-7*3A / (2p*.02m) B = 3*10-5 T F = ILBsinq F = 3A*1m*3*10-5T F = 9*10-5 N (attraction) 20 pt.

Ok, ONE Physics is Phun Question. The Chicago Blackhawks won the Stanley Cup in 2010. Before that, in what year had they last won it? 5-25 25 pt.

Ok, ONE Physics is Phun Question. The Chicago Blackhawks won the Stanley Cup in 2010. Before that, in what year had they last won it? 1961 25 pt.