They air is light and it pushes the ball around
Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
Learn more about resultant vector here: brainly.com/question/28047791
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Secondary wave is the answer
Answer:
<h2>
<em>6,142mm²</em></h2>
Explanation:
Given the dimension of a paper measured by a ruler as 7.4 cm wide and 8.3 cm long, the area of the paper is expressed using the area for calculating the area of a rectangle as shown;
Area of the piece of paper = Length * Width
Given length = 7.4cm
Length = 74mm (Since 10mm = 1cm)
Width = 8.3cm
Width (in mm) = 83mm
We converted to mm since the ruler used to measure has a division of 1mm.
Substituting the given values into the formula, we will have:
Area of the piece of paper = 74mm * 83mm
Area of the piece of paper = 6,142mm²
<em>Hence, the area of the piece of paper is 6,142mm²</em>
