On a roller coaster, riders can experience a force of up to 4 g. What is the maximum acceleration of the roller coaster?

Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

8 0

2 years ago

HELP the decay curve of twizzlers sheet!!!!!

Vinvika [58]

Ok so the first thing you need to do is im not gonna help you

7 0

2 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

2 years ago

Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m

Rufina [12.5K]

Answer:

$1.9 kg/m^3$

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

$d=\frac{42 kg}{22 m^3}=1.9 kg/m^3$

8 0

2 years ago

What do you understand by upthrust Force

Sedaia [141]

Answer:

hope you like it

Explanation:

An object that is partly, or completely, submerged experiences a greater pressure on its bottom surface than on its top surface. This causes a resultant force upwards. This force is called upthrust . The upthrust force is equal in size to the weight of the fluid displaced by the object.

Buoyancy or upthrust, is an upward force exerted by a fluid that opposes the weight of an immersed object.It is the force that pushes an object up. The upthrust, or buoyancy, keeps ships afloat. The upthrust, or buoyancy, keeps swimmers on top of the water.

7 0

3 years ago