(a) The object moves with uniform velocity from A to B.
(b) The object moves with constant velocity from B to C.
(c) The object moves with increasing velocity from C to D.
<h3>
Velocity of the object from point A to B</h3>
V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s
<h3>
Velocity of the object from point B to C</h3>
V(B to C) = (6 - 6)/(11 - 4) = 0 m/s
<h3>
Velocity of the object from point C to D</h3>
V(C to D) = (7 - 6)/(12 - 11) = 1 m/s
final velocity = 1 + 1.5 m/s = 2.5 m/s
Thus, we can conclude the following;
The object moves with uniform velocity from A to B.
The object moves with constant velocity from B to C.
The object moves with increasing velocity from C to D.
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B Ik it all just know who they’ll you
Answer:
ive once eaten octopus.. i dont recommend.. ANYWAY
i would rather shoot spaghetti out of my fingers ofc, that like free food and sneezing meatballs just sounds painful
Explanation:
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
