Hello, how to do qn 5?:)l
1 answer:
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Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!
Answer:
- 59.97 º at 18.23 mph
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph
- Vector notation:
<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph
- Vector notation:
<u>3. Addition:</u>
You add the corresponding components:
To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse:
- arctan (1.73) ≈ 59.97º