Answer:
The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.
Explanation:
Given that,
Acceleration of the car, 
Initial speed of the car, u = 24.6 m/s
Final speed of the car, v = 26.8 m/s
We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :
t = 0.84 seconds
So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.
The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
brainly.com/question/28045397
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Answer:
when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.
when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.
when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.
when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.
Explanation:
it would be at least 9.8m/s
