An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The we

The gravitational force between Royce and Earth would double by the age of 16. Newton’s law of universal gravitation states that the gravitational force is proportional to the masses involved and inversely proportional to the square of the distance between them. At age 10, Royce weighed 30 kg, and by age 16, he weighed 60 kg. Since his mass doubled from 10 to 16 years, this results in a corresponding doubling of the gravitational force.

Answer:

All observers are accurate.

Explanation:

This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.

From their distinct frames of reference, each observer's perspective is valid.

Observer A is in an inertial reference frame.

Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.

Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.

Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.

This is exemplified by observer B, who notes the car's force impacting the book's velocity.

Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.

Here is an image displaying the correct answers.

The height measures 69.68 m

Explanation:

Given data

Before striking the ground =  46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

Thus, the height is 69.68 m

A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.