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Contact [7]
2 years ago
10

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex

t, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
Physics
1 answer:
love history [14]2 years ago
8 0

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

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Answer:

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

Explanation:

  1. 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.
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  3. 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.
  4. 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}.
  5. Rearranging: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. 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.
  7. <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
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Jill puts her face in front of a convex mirror, 18 cm from the focal point of the mirror. If the focal point is located 12 cm fr
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Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

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The exercise tell us

P₂ = 1.0013 105 Pa

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      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

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2 years ago
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