Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.
Then, we find the final velocity
The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.
Answer:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:
t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
Answer:
The typical weight of a human heart is approximately 0.93 lbs.
Explanation:
Based on this,
the heart's weight constitutes about 0.5% of total body mass.
Total human weight = 185 lbs
Let the entire body weight be represented as w and the heart's weight as 
.
We aim to determine the heart's weight for a human
Using the provided information
Where, h = heart weight
w = human weight
The final weight of a human heart is 0.93 lbs.
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
Response:
a) 
, b) 
Clarification:
a) The absolute pressure at a depth of 27.5 meters is:
b) The force applied by the water is:
