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Maru [420]
3 years ago
10

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

5 kV. Find the line charge density on the power line. Wolfson, Richard. Essential University Physics, Volume 2 (p. 431). Pearson Education. Kindle Edition.
Physics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

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Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

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8 0
1 year ago
A Chinook (King) salmon (Genus Oncorynchus) can jump out of water with a speed of 6.75 m / s . If the salmon is in a stream with
Pepsi [2]

Answer:

The maximum height that the fish can jump is 2.19 m.

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).

The position and velocity vectors of the salmon at a time t, can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position of the salmon at time t.

x0 = initial horizotal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration of gravity.

Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:

vy = v0y + g · t

To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:

v0² = v0x² + v0y²

Solving for v0y:

v0y = √(v0² - v0x²)

v0y = √((6.75 m/s)² - (1.65 m/s)²)

v0y = 6.55 m/s

Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):

vy = v0y + g · t

0 = 6.55 m/s + (-9.8 m/s²) · t

-6.55 m/s / -9.8 m/s² = t

t = 0.67 s

Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:

y = y0 + v0y · t + 1/2 · g · t²

y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²

y = 2.19 m

The maximum height that the fish can jump is 2.19 m.

4 0
3 years ago
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