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aleksandr82 [10.1K]
2 years ago
5

A thin stream of water flows vertically downward. the stream bends toward a positively charged object when it is placed near it.

the positively charged object is then removed. what will happen to the same stream of water when a negatively charged object is placed near it? explain.
Physics
1 answer:
vfiekz [6]2 years ago
5 0
I dont know the answer i just want brainy points<span />
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A cylindrical specimen of brass that has a diameter listed above, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is
Irina-Kira [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The strain experienced by the specimen is 0.00116 which is option A

Explanation:

The explanation is shown on the second uploaded image

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2 years ago
Explain why it takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron fr
slava [35]
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
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2 years ago
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Imagine two people standing at placemark A and placemark E, looking at each other across the fault. Which of the following state
lord [1]

Answer:

To both observers, the land opposite them is moving to the right.

Explanation:

I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .

https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/

5 0
2 years ago
Consider the reaction data. A ⟶ products T ( K ) k ( s − 1 ) 225 0.385 525 0.635 What two points should be plotted to graphicall
lutik1710 [3]

Answer:

Plot ln K vs 1/T

(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol

Explanation:

This is an example of the Arrhenius equation:

k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = - \dfrac{E_{a}}{R}\dfrac{1}{T} + \ln A\\\\y = mx + b

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

Data:

\begin{array}{cccc}\textbf{k/s}\mathbf{^{-1}} &\mathbf{\ln k} & \textbf{T/K} & \mathbf{1/T(K^{-1})}\\0.285 & -0.9545 & 225 &0.004444\\0.635 & -0.4541 & 525 & 0.001905\\\end{array}

Calculations:

(a) Rise

Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004

(b) Run

Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹

(c) Slope

Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹

(d) Activation energy

Slope = -Eₐ/R

Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol

4 0
2 years ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
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