Weather
25/08/2016
Notes:
Atomsphere (ATM) ;
Surivival dependends on the a set of concetrated gases
The gravity compreses the atmosphere to a thin layer
composed of billions of a air molecules = ATM
higher altitude less air molecule, equals less oxygen
AT
2.97: For the purpose of doing all four parts with the least repetition of algebra, quantities will be denoted symbolically. That is, 1 2 1 2 let y1 h v0t gt , y2 h g t t0 . In this case, t0 1.00 s . Setting 2 2 2 y1 y2 0, expanding the binomial t t0 and
2.96: The time spent above ymax/2 is
1 2
the total time spent in the air, as the time is
proportional to the square root of the change in height. Therefore the ratio is 1/ 2 1 2.4. 1 1/ 2 2 1
2.95: For convenience, let the student's (constant) speed be v0 and the bus's initial position be x0. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial ve
2.94: a) The speed of any object falling a distance H h in free fall is 2 g ( H h). b) The acceleration needed to bring an object from speed v to rest over a distance h is
v2 2h 2 g ( H h) 2h g H h 1.
c)
2.93: The velocities are v A 2 t and vB 2t 3t 2 a) Since vB is zero at t = 0, car A takes the early lead. b) The cars are both at the origin at t = 0. The non-trivial solution is found by setting xA = xB, cancelling the common factor of t, and solving the
2.92: a) The time is the initial separation divided by the initial relative speed, H/v0. More precisely, if the positions of the balls are described by y1 v0 t (1 2) gt 2 , y 2 H (1 2) gt 2 , setting y1 = y2 gives H = v0t. b) The first ball will be at the
2.91: a) The final speed of the first part of the fall (free fall) is the same as the initial speed of the second part of the fall (with the Rocketeer supplying the upward acceleration), and assuming the student is a rest both at the top of the tower and
2.90: a) Suppose that Superman falls for a time t, and that the student has been falling for a time t0 before Superman's leap (in this case, t0 = 5 s). Then, the height h of the building is related to t and t0 in two different ways: 1 2 h v0 y t gt 2 1 2
2.89: a) Let +y be upward. y y0 15.0 m, t 3.25 s, a y
9.80 m s 2 , v0 y
?
20 .5 m s
y y0
v0 y t
1 2
a y t 2 gives v0 y 11.31 m s
v0 y a y t to solve for v y : v y
Use this v0y in v y
b) Find the maximum height of the can, above the point where it falls fr
2.88: a)
t fall tsound return 10 .0 s tf ts 10 .0 s
(1)
d Rock
d Sound
1 2 gt f vs t s 2
1 (9.8 m s 2 )t f2 2
(330 m s)ts
(2)
Combine (1) and (2): t f
1.16 s m h vs ts (330 )(1.16 s) 383 m s b) You would think that the rock fell for 10 s, not 8.84 s, so y
2.87: Take +y to be downward. Last 1.0 s of fall: y y0 v0 y t 1 a y t 2 gives h 4 v0 y (1.0 s) (4.9 m s 2 )(1.0 s)2 2 v0y is his speed at the start of this time interval. Motion from roof to y y 0 3h 4 : v0 y 0, v y ?
2 2 v y v0 y 2a y ( y y0 ) gives v y
2.86: a) The helicopter accelerates from rest for 10.0 s at a constant 5.0 m s 2 . It thus reaches an upward velocity of v y v0 y a y t (5.0 m s 2 )(10.0 s) 50.0 m s and a height of y 1 a y t 2 1 (5.0 m s 2 )(10.0 s)2 250 m at the moment the engine is shu
2.85: a) Let +y be upward. At ceiling, v y 0, y y0 3.0 m,a y
2 vy 2 v0 y
9.80 m s 2 . Solve for v0y.
2a y ( y
y0 ) gives v0 y
7.7 m s .
b) v y v 0 y a y t with the information from part (a) gives t 0.78 s . c) Let the first ball travel downward a distance
2.84: a) In 3.0 seconds the teacher falls a distance
1 2 1 gt (9.8 m s 2 )(9.0 s 2 ) 44.1 m 2 2 To reach her ears after 3.0 s, the sound must therefore have traveled a total distance of h (h 44.1) m 2h 44.1 m ,where h is the height of the cliff. Given 340
2.83: a) From Eq. (2.14), with v0=0,
vy 2a y ( y y0 ) 2(45.0 m s 2 )(0.640 m) 7.59 m s.
b) The height above the release point is also found from Eq. (2.14), with v 0 y 7.59 m s, v y 0 and a y g,
h
2 v0 y
2g
(7.59 m s) 2 2(9.80 m s 2 )
45 m s 2 g
2.94 m
(N
2.82: a)
b) From the speed found in part (c), the maximum height is
(111.6 m s) 2 2(9.80 m s )
2
635 m.
2
c) After the fuel is burned, the rocket has speed (40.0 m s )(2.50 s) = 100 m/s and has reached a height ( 1 2 )(40.0 m s )(2.50 s)2 = 125 m. The spe
2.81: a) The football will go an additional
v2 2g
5.00 m s 2 9.80 m s
2 2
1.27 m above the window, so
the greatest height is 13.27 m or 13.3 m to the given precision. b) The time needed to reach this height is
2(13.3 m) (9.80 m s 2 ) 1.65 s.
2.80: If the speed of the flowerpot at the top of the window is v0, height h of the window is
h vavet v0t (1 2) gt 2 , or v0 h t (1 2) gt.
The distance l from the roof to the top of the window is then 2 2 v0 (1.90 m) /(0.420 s) (1 2)(9.80 m s )(0.420 s)2
2.79: a) v
2 gh
2(9.80 m s 2 )(21.3 m)
20.4 m s ; the announcer is mistaken.
b) The required speed would be
v0 v2 2 g ( y y0 ) (25 m s)
2
2(9.80 m s 2 )( 21.3 m) 14.4 m s,
which is not possible for a leaping diver.
2.78: The elevators to the observation deck of the Sears Tower in Chicago move from the ground floor to the 103rd floor observation deck in about 70 s. Estimating a single floor to be about 3.5 m (11.5 ft), the average speed of the elevator is 103703s.5 m
2.77: Let t1 be the fall for the watermelon, and t2 be the travel time for the sound to return. The total time is T t1 t 2 2.5 s . Let y be the height of the building, then, y 1 gt12 and y vs t 2. There are three equations and three unknowns. Eliminate t2
2.76: The time needed for the egg to fall is
t
2 h 9
2(46.0 m
1.80 m)
2
(9.80 m s )
3.00 s,
and so the professor should be a distance vyt = (1.20 m s )(3.00 s) = 3.60 m.
2.75: a) The particle's velocity and position as functions of time are
v x (t ) v0 x v0 x
x(t )
t 0
t 0
( 2.00 m s ) (3.00 m s )t ) dt
2
2
3
(2.00 m s )t
3.00 m s 2
3
t2,
(0.50 m s 3 )t 3
v x (t )dt
v0 x t (1.00 m s 2 )t 2
t (v 0 x
(1.00 m s 2 )t
(0.50 m
2.74: a) From Eq. (2.17), x(t) = t t 3 = (4.00 m s )t (0.667 m s )t3. From Eq. 3
3
(2.5), the acceleration is a(t) = 2t = ( 4.00 m s )t.
3
b) The velocity is zero at t
(a = 0 at t = 0, but this is an inflection point, not an
extreme). The extreme value
2.73: a) The most direct way to find the time is to consider that the truck and the car are initially moving at the same speed, and the time of the acceleration must be that which gives a difference between the truck's position and the car's position as
2
2.72: a) There are many ways to find the result using extensive algebra, but the most straightforward way is to note that between the time the truck first passes the police car and the time the police car catches up to the truck, both the truck and the ca
2.71: a) Travelling at 20 m s , Juan is x1 37 m (20 m s)( 0.80 s) 21 m from the spreader when the brakes are applied, and the magnitude of the acceleration will be 2 v1 a 2x1 . Travelling at 25 m s , Juan is x2 37 m (25 m s)( 0.80 s) 17 m from the spreade
2.70: The position of the cars as functions of time (taking x1 = 0 at t = 0) are
x1 1 2 at , 2 x2 D v 0 t.
The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t,
1 2 at 2 v0 t D 0,
the solutions to which are
t
1 a
2 v0
2aD
v
2.69: a) xT
t
2 ( 40.0 m) (2.10 m s)
(1 2)aT t 2 , and with xT
40 .0 m , solving for the time gives
6.17 s
b) The car has moved a distance 1 2 aC 3.40 m s2 aCt x1 40.0 m 64.8 m, 2 aT 2.10 m s 2 and so the truck was initially 24.8 m in front of the car. c)