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kirill [66]
2 years ago
9

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
Physics
1 answer:
Ber [7]2 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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Answer:

75.6J

Explanation:

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2 years ago
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Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical
denis23 [38]

Given Information:

Diameter of the cylindrical tub = d = 50 cm = 0.50 m  

Acceleration = α = 3g

Required Information:

1. Rotation rate in rev/min = ω = ?

2. Tangential speed in m/s = v = ?

Answer:

1. ω = 103.5 rev/min

2. v = 2.71 m/s

Explanation:

We know that centripetal acceleration is given by

α = ω²r

Where ω is the angular speed or rotation rate and r is the radius.

The relation between diameter and radius is given by

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Since it is given that the acceleration is equal to 3g where g is the gravitational acceleration 9.81 m/s².

α = ω²r

3g = ω²r

ω² = 3g/r

ω = √(3g/r)

ω = √(3*9.81/0.25)

ω = 10.84 rad/s

To convert rad/s into rev/s divide it by 2π

ω = 10.84/2π

ω = 1.752 rev/s

To convert rev/s into rev/min multiple it by 60

ω = 1.752*60

ω = 103.5 rev/min

Therefore, the rotation rate is 103.5 rev/min

2. The tangential speed can be found using

v = ωr

Where ω is the rotation rate in rad/s and r is the radius.

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A) 5.1*10^10m B) 5.4*10^6m

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Making A subject of the formula = 2.7*10^31/ (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Since the shape is a sphere, the surface area = 4πR^2(radius of the Rigel)

R = √ (0.3261*10^23/ 4*π) = 5.1 * 10^ 10m

B) repeating the same step

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of the Procyon

Make A subject of the formula

A = 2.1*10^23/(5.67*10^-8*10^16)

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The star assumed to be a sphere;

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Answer: Cost ≈ $3,217


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