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Mademuasel [1]
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
Physics
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

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Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
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Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

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x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

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hence, the friction coefficient is 0.56

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