What can be found by counting the number of troughs per second in a wave diagram

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: Drag Force is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the Terminal Velocity.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0

3 years ago

An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan

MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

$Vertical \ velocity \ (\frac{m}{s} ) = u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) = u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) = a_{v} = 9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) = a_{h} = 0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}$

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

$u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...$

7 0

3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los

Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100

and since mass and gravity are constant so it leaves us with just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction

b) Here use the equation vf^2=vi^2+2gd

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable
we know that at maximum height or peak, the velocity is 0 so vf is 0

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.

6 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s