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kramer
3 years ago
14

help me solve this: a bicyclist in the tour de france crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farth

er, his speed is 75 km/h. what was his average acceleration (in m/s square) while riding down the mountain?
Physics
1 answer:
Lana71 [14]3 years ago
7 0
First we have to convert:
75 km / h ( * 3.6 ) = 270 m/s
18 m/s = 64.8 m/s
d = v i · t + 1/2 a · t²
v = v i + a t
-----------------------------------
4000 = 64.8 · t + 1/2 a · t²
270 = 64.8 + a t
a = ( 270 - 64.8 ) / t
a = 205.2 / t
4000 = 64.8 t + 1/2 · ( 205.2 / t ) · t²
4000 = 64.8 t + 102.6 t
4000 = 167.4 t
t = 4000 : 167.4 = 23.89 s
a = 205.2 : 23.89 = 8.58 m/s²
Answer: His average acceleration was 8.58 m/s²

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Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut M_{1} = 60.0 kg

Mass of male astronaut M_{2} = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = \frac{G*M_{1}*M_{2}}{S^{2} }

G = 6.67*10^-11 \frac{m^{3} }{kg*s^{2} }

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/M_{1} = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/M_{2} = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

a_{average} = \frac{a_{1}+a_{2}}{2} = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2)   where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0
3 years ago
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