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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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ololo11 [35]

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

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7 0
2 years ago
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How long does it take for the velocity of the rain drop to reach 99% of its terminal velocity? (assume the conditions from part
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If you think about it its part a and b 
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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

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2 years ago
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

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v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
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A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
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Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

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m=15.3\ kg

Frictional force is given by :

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F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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