Ask question

Ask question

All categories

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

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)

You might be interested in

A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o

klasskru [66]

So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

Since we know momentum and mass we can now solve.

Velocity = 264/(45+2.5)
= 5.56 m/s

5 0

2 years ago

Read 2 more answers

Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p

Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0

2 years ago

Inductive charging is used to wirelessly charge electronic devices ranging from toothbrushes to cell phones. Suppose the base un

denpristay [2]

Answer:

$1.2*10^{-6}s$

Explanation:

The problem must be addressed through the concepts of electromotive force. By Faraday's law it is defined that

$\epsilon = NA \frac{dB}{dt}$

Where

$\epsilon =$ Electromotive Force

N = Number of Loops

A = Area

B = Magnetic Field (chaging through the time)

From this equation and our values, we need to find the time, then we re-arrange the equation

$dt = NA \frac{dB}{\epsilon}$

$t = (16)(2.75*10^{-4})\frac{1.50*10^{-3}}{5.50}$

$t = 1.2*10^{-6}s$

Therefore the time required for the magnetic field to decrease to zero from its maximum value is $1.2*10^{-6}s$

8 0

2 years ago

A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the

lisabon 2012 [21]

Answer:

We are given x= bt +ct²

So

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

B.

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

3 0

2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago