Use formula I provided plus don’t copy from chegg please explain me step by step would be very helpful to understand concept 6,7,8 6. A puck on a frictionless air hockey table has a mass of 0.5 kg and is attached to a cord passing through a hole in the surface as in the figure.The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 0.40 rev/s. What is the kinetic energy of the puck? answer is 6.3 J 7.

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ISBN:9781305952300
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Use formula I provided plus don’t copy from chegg please explain me step by step would be very helpful to understand concept 6,7,8 6. A puck on a frictionless air hockey table has a mass of 0.5 kg and is attached to a cord passing through a hole in the surface as in the figure.The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 0.40 rev/s. What is the kinetic energy of the puck? answer is 6.3 J 7. A 2-kg object moving to the right with velocity 5 m/s collides with a 3-kg object which is initially at rest. After the collision, the 3-kg object has a velocity vector v = i – j. What is the speed of the 2-kg object after the collision? answer is 3.8m/s 8. If 15 J of potential energy are stored in a spring which has been compressed by 2 cm from its relaxed length, how much force will this spring exert if it is stretched by 4 cm from its relaxed length? answer is 3000n
The exam is closed book and closed notes.
v2
Circular motion: a =
W
Pavr =
At
Weight: Fg= mg,;
g = 9.8 m/s?;
Fnet = ma;
1
-m v²;
Potential energy: Ug = mgy
E = Er
E = U+K
Kinetic energy: K =
y2
- = o'r;
Rotational motion:
1 rev = 2n rad;
v = o r;
a αr;
a, =
r
r
1
@ = 00 + at;
0 = 0o t + -at2;
2θα -ω- ω02
K=-Io
T=r x F;
T=rFsino;
1
1
1
Icy= mR2
2
+ Rin
Στ-Ια;
Ipoint mass = mr?
Idisk =
mR
out
1
Ihoop = mR?
Irod(center) =
mL²
Irod(end)
mL²
2
Iball = =mR
12
Ishell =
mR?
I= Icom + MD?
1
K=.
1o,
1 Jo.?
work: W=t 0;
dW
P =
2
W=
2
P.
dt
Rolling:
2
-Io
1
m Vcom
Vcom = Ro
K =
2
T= f,R
Fs,max = HsFn
Incline: F= mgsin0 Fg mgcos0
Angular momentum:
Lpoint mass =m rxv
L= mrvsin 0;
L=m (r¡Vy - ryVx)k
L=Io
stae wil
LI = L
I 01 = I 202
m1X1+m2x2
X com =
mıy1+m2y2
m¡ +m2
Y com =
m1+m2
Transcribed Image Text:The exam is closed book and closed notes. v2 Circular motion: a = W Pavr = At Weight: Fg= mg,; g = 9.8 m/s?; Fnet = ma; 1 -m v²; Potential energy: Ug = mgy E = Er E = U+K Kinetic energy: K = y2 - = o'r; Rotational motion: 1 rev = 2n rad; v = o r; a αr; a, = r r 1 @ = 00 + at; 0 = 0o t + -at2; 2θα -ω- ω02 K=-Io T=r x F; T=rFsino; 1 1 1 Icy= mR2 2 + Rin Στ-Ια; Ipoint mass = mr? Idisk = mR out 1 Ihoop = mR? Irod(center) = mL² Irod(end) mL² 2 Iball = =mR 12 Ishell = mR? I= Icom + MD? 1 K=. 1o, 1 Jo.? work: W=t 0; dW P = 2 W= 2 P. dt Rolling: 2 -Io 1 m Vcom Vcom = Ro K = 2 T= f,R Fs,max = HsFn Incline: F= mgsin0 Fg mgcos0 Angular momentum: Lpoint mass =m rxv L= mrvsin 0; L=m (r¡Vy - ryVx)k L=Io stae wil LI = L I 01 = I 202 m1X1+m2x2 X com = mıy1+m2y2 m¡ +m2 Y com = m1+m2
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