Ask question

Ask question

All categories

2 years ago

6

Mary and her younger brother Alex decide to ride the 17 -foot-diameter carousel at the State Fair. Mary sits on one of the horse

s in the outer section at a distance of 8 feet from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 5 feet from the center.
(a) What is Mary's angular speed ωM compared to that of Alex's angular speed ωA ? Give your answer as a multiple of ωA.ωM=ωA
(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA? Give your answer as a multiple of vA.vM=vA

1 answer:

Elodia [21]2 years ago

7 0

Answer:

(a) ωM=ωA

(b) vA = (0.625)vM

Explanation:

The tangential velocity of the body is calculated by the following formula:

v = ω*R Formula (1)

where:

v is the tangential speed of the body in feet over second (ft/s)

ω is the angular speed in radians over second (rad/s)

R is the radius of the the body in the circular path in meters (m)

(a) What is Mary's angular speed ωM compared to that of Alex's angular speed ωA

The angular speed of Mary and Alex is determined by the angular speed of the carousel ( ωC),therefore it is the same for both:

ωM=ωC , ωA=ωC

ωM=ωA

(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA?

Data:

RM = 8 ft : Mary radio

RA = 5 ft : Alex radio

We apply the formula (1)

v = ω*R

vM = ωC *8 : Mary's tangential speed

vA = ωC *5 : Alex's tangential speed

We establish the relationship between vA and vM like this:

vA/vM = (ωC*5) / (ωC*8)

We cancel ωC

vA = (5/8)vM

vA = (0.625)vM

You might be interested in

sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he

podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0

2 years ago

The rock starts as a solid and then..

alina1380 [7]

Answer:

they stay a solid

Explanation:

breaks apart do to whether such as erosion

7 0

3 years ago

Which formula correctly expresses the property density?

Olenka [21]

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

6 0

2 years ago

Which is a consequence of the third law of thermodynamics Energy is not always conserved. Engines cannot discharge waste heat. H

miss Akunina [59]

Heat engines are less than 100% efficient because absolute zero cannot be reached

8 0

3 years ago

Read 2 more answers

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago