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V125BC [204]
3 years ago
13

Which box plot represents the data? 30, 35, 25, 5, 5, 25, 40, 45, 50, 10, 15, 40

Mathematics
1 answer:
Sindrei [870]3 years ago
7 0
It's the first box plot
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What is the are of the figure below​
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You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 15 km/hr and your
Viefleur [7K]

Answer:

a)

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin  (2πt))

b) the  foot have to be 3.32 rev/sec faster

Step-by-step explanation:

Given that:

the speed of the bike = 15 km/hr = 15 × 1000/3600 (m/sec) = 4.167 m/sec

radius of the circle when the foot moves = 20 cm = 0.2 m

radius of the circle above the ground = 30 cm = 0.3 m

Let assume that:

x(t)  should represent the vector along the horizontal moment

y(t) should be the vector along the vertical moment

The initial component will be ( 0, 0.3)

We know that the radius of the circle is given as 0.2 m, So the vector of the circle can be written as (0.2 cos t , 0.2 sin t )

Also, the foot makes one revolution in a second, definitely the frequency of the revolution = 1 and the vector for the circle is ( 0.2 cos (2πt), -0.2 sin  (2πt)), due to the fact that the foot moves clockwise.

Thus, adding all the component together ; we have:

(x(t), y(t)) = (0,0.3)+(4.167 t , 0)+(0.2 cos (2πt), -0.2 sin  (2πt))

(x(t), y(t)) = (4.167 t + 0.2 cos (2πt), 0.3 - 0.2 sin  (2πt))

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x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin  (2πt))

b)

The linear speed of rotation is :

15km/hr = 15 × 100, 000/3600 (cm/sec)

             = 416.7 cm/sec

The rotational frequency is :

= 416.7/2πr

= 416.7/2(3.14 × 20)

= 3.32 rev/sec

Hence, the  foot have to be 3.32 rev/sec faster in rotating if an observer standing at the side of the road sees the light moving backward.

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