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3 years ago

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Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P

osition 1.8, 5.0, 0 m 0.5, 6.9, 0 m −0.4, 9.5, 0 m (a) Between 6.6 s and 6.9 s, what was the bee's average velocity? Be careful with signs. vavg, a = 7.6739 (b) Between 6.6 s and 7.2 s, what was the bee's average velocity? Be careful with signs. vavg, b = 4.58557 (c) Of the two average velocities you calculated, which is the best estimate of the bee's instantaneous velocity at time 6.6 s? (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s? Δr = m

1 answer:

Ber [7]3 years ago

5 0

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

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Kaushik had three pendulum bobs made of steel of the same length but different masses of 10 g, 50 g and 100 g respectively. When

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In order to answer this question, we need to analyze the parameters, upon which the time period of a pendulum depends. We know that the time of a pendulum is given by the following formula:

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The formula clearly shows that the time period of the pendulum depends only upon the length of pendulum and value of g. And the time period of a pendulum does not depend upon the mass of the bob. Hence, the time period for each of the three pendulums will remain same. So, the correct option will be:

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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m

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Answer:

a) total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

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As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is d1=0.5*L

By Steiner theorem

for the rod we get $J_1'=J_1 + m_1\times d_1^2$

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for the sphere we get $J_2' = J_2 + m_2\times d_2^2$

$J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2$

And the total moment of inertia for the first case is

$J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2$

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The torque for system is given as

$M = F\times d\times sin(a)$

where a is angle between Force and distance d

and where d represent distance from rotating axis.

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M=476*2.44 = 1161.44 Nm

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The maximum speed of the donkey is 10.72m/s

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We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

a is the acceleration of the donkey

t is the time taken in seconds

v = u + at

v= 0 + 1.6 x 6.7

= 10.72 m/s

Hence the donkey obtains the speed of 10.72 m/s

For further reference:

brainly.com/question/24478168?referrer=searchResults

#SPJ9

3 0

1 year ago