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Nutka1998 [239]
4 years ago
7

During one season of racing at the Talladega Superspeedway, the mean speed of the cars racing there was found to be 158.9 mph wi

th a standard deviation of 10.0 mph. What speed represents the 33rd percentile for speeds of race cars at Talladega? Assume that the racing speeds are normally distributed.
Physics
1 answer:
Scorpion4ik [409]4 years ago
3 0

Answer:

154.5  mph

Explanation:

mean: 158.9 mph, SD: 10.0 mph

Need to convert the mean and SD into normal distribution.

We know P(X<z) = 0.33

From the normal distribution table,  P(X<z) = 0.33, z value is -0.44

P(X< -0.44) = 0.33

we know \frac{x-mean}{SD} = Z

\frac{x-158.9}{10} = -0.44\\x=154.5

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Answer:

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(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.

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3 years ago
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Answer:

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Explanation:

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