PHYS2A HW#4, Ch.6
3) A 3700 kg barge is pulled by 2 mules. Tension in each rope is 1100 N, @ 25o. Find force of water if (a) constant velocity and (b)
a = 0.16 m/s2.
2T cos 25o Fres = max, so (a), 2(1100 N)*cos 25o = 1990 N, and (b), 1990 N (3700 kg)
LECTURE 3 (M 2)
JOURNEY TO THE CENTER
OF THE EARTH
If you sliced through
the Earth, as if it
were a round egg,
what would you see?
Earth and its Neighborhood
What is the Earth made of?
How do we know t
car47872_ch23_964-989 11/15/06 16:33PM Page 980
The rst evidence that 1,4-dehydrobenzene could be generated as a reactive intermediate came in 1972 from studies of the thermal isomerization of the deu
SYSTEMS OF PARTICLES
22. A rowboat with a mass of 94 kg carries a 61-kg rower and 34 kg of water that has accumulated in the
bottom. Its drifting forward at 0.84 m/s when the rower scoops up 11 kg of water in a bucket of
CHAPTER 4 MOTION IN MORE THAN ONE DIMENSION
An object is moving initially in the x direction at 4.5 m/s , when an acceleration is applied in the y direction for a
period of 18 s. If it moves equal distances in the x and y directions during this
4) Airliner velocity
240i m/s + (0.38i + 0.72j) m/s2 * (24 s) = (269i + 17.3j) m/s, angle of deflection is arctan(17.3/269) = 3.67 degrees
6) Airplane down a runway, express planes velocity and position at time t = 30s
acceleration @ NE, so 45 degrees, so
PHYS2A, Ch. 14, HW#8
2) A body is subject to the forces F1 = 2i + 2j N at (2, 0), F2 = -2i 3j N at (-1, 0), and F3 = 1j N at (-7, 1). (a) Show that net force is 0. (b) Show that net torque
about the origin is zero.
(a) Fi = (2i + 2j 2i 3j + j) N = 0,
Easy (#4 and #42 nothing special)
1) A mass weighs twice as much on a planet the mass of earth than earth itself. Find radius.
g = GM/R2. Therefore, (gp/ge) = (Mp/ME)(RE/Rp)2 = 2, (Mp/ME) = 1, so Rp = RE/2
3) To what fraction of its current radius w
PHYS2A HW#6, 10, 11
1) A 28 kg child sits on a 3.5 m seesaw. Where should 65 kg father be so that center of mass is at center of the seesaw?
Center of mass @ origin, xcm = 0 = mcxc + mfxf, xc = -(3.5 m)/2. Hence, xf = -mcxc/mf = 75.4 cm from center.
Phys2A, HW Ch. 5
4) Force of the seatbelt on the person is equal and opposite to the force of the person on the seat belt.
F = ma, F = (60 kg)(30.56 m/s)/(0.14 s) = 13095 N, answer is -13095 N, since its the seat belt acting on the person (opposite d
PHYS2A HW#7, Ch 12,13
5) A wheel turns through 2.0 revs while being accelerated from rest at 18 rpm/s. (a) Find angular speed and (b) time to make the 2 revs
(a) (12-11) wf = sqrt(w2O + 2a(f - O) = sqrt(0 + 2(18 rev * 60/min2)(2 rev) = 65.7 rpm
Phys2A Ch 7&8, Hw#5
3) Crane lifts 650 kg mass up 23 m. Swing it east 18 m. Find work.
Only account for the vertical. W = mgy = (650 kg)(9.80 m/s2)(23 m) = 147 KJ
4) You lift 45 kg mass up 2.5 m. (a) Find work. (b) Find work holding it up in place fo
Chemistry 140C Spring 2015 (K. Albizati)
1. Chemistry of Carboxylic Acids
a. General Structure, Reactivity and Nomenclature
b. Acidity and Basicity
c. Methods of Synthesis
d. Reactions of Carboxylic Acids
6. Heterocyclic Chemistry