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3 years ago

15

If the vertical posts supporting these wires divide the lines into straight 19.0 m segments, what magnetic force does each segme

nt exert on the other?

1 answer:

erica [24]3 years ago

5 0

Answer:

Complete question is:

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each other 38.0 cm apart.

Part A.) If the vertical posts supporting these wires divide the lines into straight 20.0m segments, what magnetic force does each segment exert on the other? (answer in F= ? N)

Part B.) Is the force found in part (A) attractive or repulsive?

Answer: 0.024 N attraction force is being applied.

Explanation:

See attached picture.

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Solve for (b) how many revolutions it takes for the cd to reach its maximum angular velocity in 1.36s?

Digiron [165]

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

$\omega = 30 rad/s$

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

$\theta = (\frac{\omega_0 + \omega}{2})t$

where:

$\theta$ is the angular displacement of the cd during the time interval t

$\omega_0$ is the initial angular velocity of the cd

$\omega$ is the final angular velocity

Here we have:

t = 1.36 s

$\omega_0 = 0$ (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

$\theta=(\frac{0+30}{2})(1.36)=20.4 rad$

And since $1 rev = 2 \pi rad$ , we can convert into number of revolutions completed:

$\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev$

Learn more about angular motion:

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8 0

3 years ago

: Write your reflection after reading the statement below.

max2010maxim [7]

It was not the first time that a person who was in a few weeks ago, I

7 0

3 years ago

One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is movi

erastova [34]

Answer:

(a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

Speed = 81 mph = 36.2 m/s

Distance = 81 cm

(a). We need top calculate the ball's centripetal acceleration just before it is released

Using formula of centripetal acceleration

$a=\dfrac{v^2}{r}$

Where, v = speed

r = radius

Put the value into the formula

$a=\dfrac{(36.2)^2}{81\times10^{-2}}$

$a=1617.82\ m/s^2$

$a=16.17\times10^{2}\ m/s^2$

(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

Using formula of force

$F=\dfrac{mv^2}{r}$

Put the value into the formula

$F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}$

$F=232.9\ N$

Hence, (a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

4 0

3 years ago

A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

3 years ago

Read 2 more answers

Help with 1 2 and 3 please

geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$

We are looking for L2 so we can isolate using algebra to get:

$\frac{A2(\frac{P1L1}{A1}) }{P2} = L2$

If you fill in those values you get 0.0205

or 2.05 cm

6 0

3 years ago