Physics URGENT Please Help Calculus? - long travel sand cars
1. A hot air balloon is rising upward with a constant speed of 6.5 m / s. When the balloon basket is 20m high and a bag of sand is released bound in the basket
* (A) After he released from the basket, how long is the bag of sand in the air before it hits the ground? (Caution!)
* (B) What is the climax of the bag is in its fall to the ground
2. A person who is on a bridge and throw a ball with an initial vertical velocity of 11.2 million / s straight upward. (Note: The size of the acceleration due to gravity has magnitude 9.81 m/s2.)
* (A) What is on the bridge of the ball?
* (B) How long does it take to get the ball to get there?
* (C) What is your height 2.0s after the person who threw it?
* (D) As the ball with a speed of 5.0 m / s is it? It is a difficult question, therefore, the interpretation of the results of his physical. (Note: The physical interpretation is the most important part of that question! A correct numerical answer with false [or] physical interpretation will receive 0 points!)
3. To test the case of a collision, driving a car to 100km/hr length 1.5 m (approx. 62 mph), hit a concrete wall buildings. The car stops when the length is reduced by half. What is the acceleration? (Use SI units.)
4. A race car is at rest and accelerated to 7,40 m/s2. To what extent it has traveled 3.00?
Tuesday, January 19, 2010
Long Travel Sand Cars Physics URGENT Please Help Calculus?
3 comments:
1. a)
V (2) ^ 2 = V (1) ^ 2 +2 ad
0 = (6.5) ^ 2 +2 (-9.81) d
-42.25 =- 19.62d
d (a) = 2.153m
D (total) = 20 2153
D (total) = vt +1 / 2AT ^ 2
= 22.153 (0) T 1 / 2 (9.81) t ^ 2
4.905T ^ 2 = 22.153
4.516 = t ^ 2
t = 2.125
b) d = 22.153m
2. a) v (2) ^ 2 = v (1) ^ 2 +2 ad
(0) ^ 2 = (11.2) ^ 2 +2 (-9.81) d
125.44 = 19.62d
d = 6.393
b) d = vt +1 / 2AT ^ 2
6,393-4.905 t = 11.2t ^ 2
0 =- 4.905t ^ 2 11.2 t-6, 393
(Solution Formula reply)
Ok, that everything that I very tough spot right on your computer
4. d = vt +1 / 2AT ^ 2
= (0) (3) 1 / 2 (7.4) (3) ^ 2
= 33.3 M
That was too easy to make = P
1. a)
V (2) ^ 2 = V (1) ^ 2 +2 ad
0 = (6.5) ^ 2 +2 (-9.81) d
-42.25 =- 19.62d
d (a) = 2.153m
D (total) = 20 2153
D (total) = vt +1 / 2AT ^ 2
= 22.153 (0) T 1 / 2 (9.81) t ^ 2
4.905T ^ 2 = 22.153
4.516 = t ^ 2
t = 2.125
b) d = 22.153m
2. a) v (2) ^ 2 = v (1) ^ 2 +2 ad
(0) ^ 2 = (11.2) ^ 2 +2 (-9.81) d
125.44 = 19.62d
d = 6.393
b) d = vt +1 / 2AT ^ 2
6,393-4.905 t = 11.2t ^ 2
0 =- 4.905t ^ 2 11.2 t-6, 393
(Solution Formula reply)
Ok, that everything that I very tough spot right on your computer
4. d = vt +1 / 2AT ^ 2
= (0) (3) 1 / 2 (7.4) (3) ^ 2
= 33.3 M
That was too easy to make = P
1. a) d = .5 at ^ 2
2 sec.
1. b) 20 m (driving at a constant speed, the acceleration is not up or force acting on it. He will not go upstairs to be released)
2. a) d = .5 at ^ 2
6,39 m
2. b) v = en
1.14 s
2. c) A 2 s my journey to and fro ... and up to 86 s. d = .5 at ^ 2
3.63 meter
2. d) the time for the purchase of 11.2 m / s to 5 m / s = 6.2 for the use, t = 63 s
then use the d = .5 ^ 2, d = 1.96 m (which has a speed of 5 m / s to 5 m / s down so that) at this point
3. Use v ^ 2 = Vo ^ 2 +2 AD (acceleration is negative)
1111.11 m / s ^ 2
4. Use at D = .5 ^ 2
33.3 m
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