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faltersainse [42]
2 years ago
5

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte

rval, the wheel rotates through 70.4 rad. what is the angular speed of the wheel at the end of the 3.80-s interval
Physics
1 answer:
ddd [48]2 years ago
5 0

<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


</span>

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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
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Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

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T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

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2 years ago
Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
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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

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