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Mademuasel [1]
3 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]3 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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ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

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To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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Answer:

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\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

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