a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

The height measures 69.68 m

Explanation:

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

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

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m

Answer:

The distinction between the locations of the two events as perceived in S is calculated to be 4.42 x 10^8 m

Explanation:

The time dilation concept has been utilized to address this question, as detailed in the attachment.