
Tension, Omega, and Trigonometric Relationships
Interactive Video
•
Physics
•
11th - 12th Grade
•
Hard
Lucas Foster
FREE Resource
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10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the relationship between velocity, radius, and omega?
Omega is the sum of velocity and radius.
Omega is the difference between velocity and radius.
Omega is the ratio of velocity to radius.
Omega is the product of velocity and radius.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why does the speaker prefer using omega instead of omega squared initially?
To simplify the calculation of velocity.
Because omega squared is more complex.
Because omega squared is not needed.
To avoid confusion with too many steps.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How is tension calculated using mass, length, and omega squared?
Tension equals mass times length times omega squared.
Tension equals mass divided by length times omega squared.
Tension equals mass times length times omega.
Tension equals mass times length divided by omega squared.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the significance of evaluating tension in terms of g?
It eliminates the need for omega.
It provides a precise answer.
It makes the calculation more complex.
It simplifies the calculation.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How is theta found using trigonometric relationships?
Theta is the sine of the opposite side over the hypotenuse.
Theta is the tangent of the opposite side over the adjacent side.
Theta is the cosine of the opposite side over the hypotenuse.
Theta is the cotangent of the opposite side over the hypotenuse.
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why does the speaker argue for using radians over degrees?
Radians are easier to visualize.
Degrees are not used in physics.
Radians are simpler to calculate.
Radians are the natural unit for calculus.
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the alternative method to find omega?
Using the derivative of velocity.
Using the sum of velocity and radius.
Using the circumference of the base circle.
Using the quadratic equation.
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