What is the angular displacement of the wheel between t = 5 s and t = 15 s?
1 answer:
The question seems incomplete. The complete text is:
a)What is the angular displacement of the wheel between t = 5 s and t = 15 s
b)What is the angular velocity of the wheel at 15 s
And it refers to the attached figure.
a) 25 rad
The graph shown represent the angular position of the wheel at different times.
Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.
At , the angular position from the graph is
At , the angular position from the graph is
Therefore, the angular displacement is
2) -5.0 rad/s
For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.
Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.
By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:
This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s
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<h2>Answer:</h2>
(a) standard deviation = σ = 4.9996
(b) variance = σ² = 24.996
<h2>Explanation:</h2><h2 /><em>Given frequency table (find attached as Table 1);</em>
<u></u>
(a) To find the sample standard deviation and sample variance, follow these steps;
<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>
c = (lower bound + upper bound) / 2
=> c = (6.51 + 8.50) / 2 = 7.505
=> c = (8.51 + 10.50) / 2 = 9.505
=> c = (10.51 + 12.50) / 2 = 11.505
=> c = (12.51 + 14.50) / 2 = 13.505
=> c = (14.51 + 16.50) / 2 = 15.505
<em>So the new table becomes (find attached as Table 2);</em>
<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>
n = 50+18+42+20+46
n = 176
<em>iii. Calculate the mean (M)</em>
This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).
M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176
M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176
M = [2012.88] / 176
M = 11.44
<em>iv. Find the variance (σ²);</em>
The variance is calculated using the following formula
σ² = [Σ(f x c²) - (n x M²)] / (n - 1) ------------(i)
Where;
f = frequency of each boundary data point
<em>=> Let's first calculate </em>Σ(f x c²).
This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)
Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]
Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]
Σ(f x c²) = 24708.1244
<em>=> Now calculate (n x M²)</em>
n x M² = 176 x 11.44²
n x M² = 23033.7536
<em>=> Now substitute these values into equation (i) to calculate the variance</em>
σ² = [Σ(f x c²) - (n x M²)] / (n - 1)
σ² = [24708.1244 - 23033.7536] / (176 - 1)
σ² = [4374.3708] / (175)
σ² = 24.996
Therefore, the variance is 24.996
<em>v. Find the standard deviation (σ)</em>
The standard deviation is the square root of the variance. i.e
σ = √σ²
σ = √24.996
σ = 4.9996
Therefore, the standard deviation is 4.9996
Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:
II) using these two equations after substitution v₀=0; v=30 and L=45: