The tension has to hold the part of the weight in the direction of the string:
T = mg*cos(theta)
Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah
The answer is,
<u>Farthest from the axis of rotation</u>
Please rate <u>Brainliest</u> (:
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
Answer:
the assessment is due at 11 you only have 1 hour do it fasst
Explanation:
the assessment is due at 11 you only have 1 hour do it fasst
