Among the statements, B) An object at rest has an instantaneous acceleration of zero, is correct. This is because when an object is at rest, it isn't moving, and hence, no change in velocity and no acceleration. Other statements do not hold true in all cases.
The statement that is true among the options provided is B) An object at rest has an instantaneous acceleration of zero. When an object is at rest, it means it's not moving, and therefore, it doesn't have any acceleration. Acceleration is the rate at which an object changes its velocity. Thus, if an object is not moving, there is no change in velocity and hence, no acceleration. As for the other statements, A, C, and D, they do not hold true for all cases. An object can be accelerating while moving in a straight line (changing speed but not direction). Also, instantaneous acceleration isn't always changing - it can remain constant. Lastly, an object that is accelerating isn't always changing speed, as it could be moving in a circle at a constant speed.
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The only fully accurate statement is that an object at rest has an instantaneous acceleration of zero, as no change in velocity indicates no acceleration. Other statements have instances where they might not always be true in the field of physics.
According to the fundamentals of Physics, only option B is fully correct: An object at rest has an instantaneous acceleration of zero. This is because an object at rest will stay at rest, with no change in velocity that would constitute acceleration.
Option A) is not entirely correct. An object that is accelerating may change direction, but not always. For example, a car can accelerate while driving straight.
Option C) is also not absolute. Instantaneous acceleration could be changing, but it could also be a constant rate if an object is moving with a uniform acceleration.
Finally, option D) is like A, is not always true: an object that is accelerating doesn't necessarily have to be changing speed. It could also be changing direction while maintaining the same speed.
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Answer:
v = 54 m/s
Explanation:
Given,
The maximum height of the flight of golf ball, h = 150 m
The velocity at height h, u = 0
The velocity of the golf ball right before it hits the ground, v = ?
Using the III equations of motion
v² = u² + 2gh
Substituting the given values in the above equation,
v² = 0 + 2 x 9.8 x 150 m
= 2940
v = 54 m/s
Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s
The floor's reaction force back up on the boy= 300N
His WEIGHT IS EVENLY DISTRIBUTED OVER EACH FOOT
What is the net force on the boy?
Speed of Dayshawn travelling towards his home is 12 m/s
Speed of her mom towards his school is 5 m/s
They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home
Now let say they both meet after "t" time when they starts motion
so we can write the total distance between school and home as
here d = 6492 m
= speed of dayshawn
= speed of his mom
now by solving the above equation
so they will meet after 381.9 s from start which will be 3.36 minutes from there start
Also at this time the distance covered by her mom will be
so they will meet at distance 1909.4 m from their home