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katen-ka-za [31]
3 years ago
9

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

charge density 3.00 uC/m.. . What is the kinetic energy of the sphere when it is 4.50cm from the line of charge if the only force on it is the force exerted by the line of charge?
Physics
1 answer:
JulijaS [17]3 years ago
5 0
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
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A feather is dropped on the moon from the height of 1.4m. The acceleration of gravity on the moon is 1.67ms-1. Determine the tim
Zinaida [17]

1.3s

Explanation:

Given parameters:

Height = 1.4m

Gravity on moon = 1.67ms⁻¹

Unknown:

Time for feather to fall = ?

Solution:

To solve this problem, we are going to use one of the motion equation that relates time, gravity and height.

    H = ut + \frac{1}{2} g t^{2}

Sine the body was dropped from rest, initial velocity is zero;

 H = height

  u = initial velocity

  t = time

  g = acceleration due to gravity

since u = 0;

H = \frac{1}{2} g t^{2}

 1.4 = \frac{1}{2} x 1.67 x t²

  t = 1.3s

learn more:

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8 0
3 years ago
When responding to sound, the human eardrum vibrates about its equilibrium position. suppose an eardrum is vibrating with an amp
vodomira [7]
745.92 hz (b) 
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5 0
3 years ago
WhAT is the change IN THE entropy of 2.0kg of h2o molecules when transform at a constant pressure of 1 atm from water at 100 deg
Fynjy0 [20]

Answer:

The entropy change is 45.2 kJ/K.

Explanation:

mass of water at 100 C = 2 kg

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Heat is

H = m L

H = 2 x 2260 = 4520 kJ

Entropy is given by

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3 years ago
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike th
alexira [117]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

4 0
3 years ago
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Ray Of Light [21]

Answer:

At the closest point

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We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

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In this problem, we have a comet orbiting around the Sun:

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- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: A\propto vr, therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster (A\propto vr, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

7 0
3 years ago
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