In the picture below, label which particle is the solute and which is the solvent

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago

Juan lives 100 m away from Bill.What is Juan's averege speed if he reaches Bill's home in 50 s?

Tom [10]

Juan lives a hundred miles away from Bill then the average speed that he reaches Bill's Home in 50 seconds means that Juan lives 50 seconds away from bill because 50 + 50 seconds equals 100 seconds so Juansorry my bad Juan lives 50 miles away from Belle but he probably how I must have run to get to bill in 50 seconds .

7 0

3 years ago

The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r

Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity. Therefore:

marble > manhole cover > basketball > wedding ring

7 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago

Which of the following is true about a planet orbiting a star in uniform circular

Mnenie [13.5K]

The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.

A. The velocity vector of the planet points toward the center of the circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.

7 0

3 years ago

In the picture below, label which particle is the solute and which is the solvent ​

In the picture below, label which particle is the solute and which is the solvent