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Vesnalui [34]
1 year ago
10

A carnival ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

ion of a passenger at his lowest point during the ride?.
Physics
1 answer:
MaRussiya [10]1 year ago
5 0

The acceleration of the person is 4.108 m/s^2.

What is acceleration?

The acceleration is the rate of change in the velocity in a unit of time.

Angular velocity: The change in angular displacement in a unit of time is called angular velocity.

Tangential velocity: The tangential velocity can be defined as the velocity of an object which is perpendicular to the radius in the rotational motion.

The relation between angular velocity ω, tangential velocity v, and radius r from the axis when the radius is perpendicular to the tangential velocity is,

v=ω*r

Given r=15 m, and ω=5 turns/minutes, substitute these values in the above formula.

Note: 1 turn/minute = 2π/60 rad/s.

v=(5 turns/minutes)*15 m

v=(5*2π/60 rad/s)* 15 m

v=7.85 m/s

Since the motion is circular, and the person is at the lowest point of the wheel, so the acceleration due to gravity will have no effect as it is perpendicular to tangential velocity here. The acceleration a for rotational motion is given by,

a=v^2/r

Substitute v=7.85 m/s, and r=15 m in this equation and solve it.

a=(2.094)^2/(15)

a=4.108 m/s^2

Learn more about acceleration here:

brainly.com/question/12550364

#SPJ4

You might be interested in
It takes 6 sec for a stone to fall the the bottom of a mineshaft. calculate the
Stels [109]
Ok so this is simple projectile motion problem.

if we have an object falling in free fall it is subject to gravity of -9.80m/s^2

so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m 

so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so 
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground


hope this helps you! Thanks!!

8 0
3 years ago
Find the net force and acceleration. 15 points. Will give brainliest!
gladu [14]

Answer:

\boxed{F_{net} = 28.7 \ N}

\boxed{a = 2.1 \ m/s^2}

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

F_{k} = (micro)_{k}mg

Where (micro)_{k} is the coefficient of friction and m = 13.6 kg

F_{k} = (0.16)(13.6)(9.8)\\

F_{k} = 21.32 \ N

<u><em>Now, Finding the net force:</em></u>

F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\

F_{net} = 28.7 \ N

<u><em>Finding Acceleration:</em></u>

a = \frac{F_{net}}{m}

a = \frac{28.7}{13.6}

a = 2.1 \ m/s^2

8 0
3 years ago
Carla holds a ball 1.5 m above the ground. Daniel, leaning out of a car window, also holds a ball 1.5 m above the ground. Daniel
spayn [35]

Answer:

Carla

Explination: As Daniel's ball is dropped from the car moving at 40 mph in a horizontal direction, at the time the ball is dropped it is also moving at 40 mph in a horizontal direction due to inertia, a property of mass causing resistance to change, Daniel's ball will continue to move in a horizontal direction even after being dropped along with falling due to gravity. Daniel's ball will then fall in a projectile motion curve of sorts which will cause an overall velocity to not be straight down causing it not to fall to the ground as quickly as Carla's ball.

Sorry for the long explanation

8 0
3 years ago
Read 2 more answers
This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi
Rufina [12.5K]

Answer:

 N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

let's calculate

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0
3 years ago
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

3 0
3 years ago
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