Which of the following correctly describes a solenoid? A) The magnetic field around a solenoid is unipolar.
B) The field around a solenoid is termed a bar magnet.
C) A solenoid is created by winding current carrying wire into a coil.
D) Increasing the current in the wires will reduce the magnetic field outside of the solenoid.

Answers

Answer 1
Answer:

Answer;

C) A solenoid is created by winding current carrying wire into a coil.

Explanation;

-A solenoid is a coil of insulated or enameled wire wound on a rod-shaped form made of solid iron, solid steel, or powdered iron.

-Solenoids have a wide range of applications, for instance they are used as electromagnets, as inductors in electronic circuits, and as miniature wireless receiving antennas.

Answer 2
Answer: Option C describes a solenoid

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with what minimum speed must you toss a 160 g ball straight up to just touch the 13-m-high roof of the gymnasium if you release the ball 1.4 m above the ground? solve this problem using energy.

Answers

energy at top = m*g*height at top from release point
=0.16*9.81*11.6
=18.21J
At release kinetic energy= Gravtiational energy at top
1/2*0.16*v^2=18.21J
v^2=227.625
v=15.10m/s

Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved, then what is the new acceleration of the sled?(show how you solved it)

Answers


Start with Newton's 2nd law of motion:

                             Force = (mass) x (acceleration) .

Since we're going to be talking about acceleration,
lets divide each side by (mass):

                             Acceleration = (force) / (mass) .

OK.  We start out with a certain acceleration 'A₀'.  It's the result of
a certain force 'F₀' and a certain mass 'M₀'.  (I used the little subscripts " ₀ "
to show that these are the originals, before any changes.)

             Original acceleration = (Original force) / (original mass)

                              A₀             =          F₀            /        M₀ .

Now you want to triple the force and cut the mass in half: 

                     New acceleration  A₁ = (3 F₀) / (1/2 M₀) .

Divide each side by 3:           A₁ / 3  =  F₀ / (1/2 M₀) .

Multiply each side by  1/2 :   (1/2 A₁) / 3  =  F₀/M₀

                                                   A₁ / 6  =  F₀/M₀

Take a look at the right side of that equation . . .  F₀/M₀ .
That's just the original acceleration  A₀ .
So now, after the change, we have                 A₁ / 6  =  A₀ .

You asked "What is the new acceleration ?"
OK.  Multiply each side by  6  :                        A₁ = 6 A₀ .

Whatever the original acceleration was, the
new acceleration is 6 times as much.

If it was originally  2 m/s², then after the change, it becomes  (6 x 2)  =  12 m/s² .

Final answer:

Using Newton's second law of motion, if the net force on a sled is tripled and the mass halved, the new acceleration would be 12 m/s².

Explanation:

The subject of this question is derived from the principle of Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be represented as: F=ma (Force = mass * acceleration).

Given that the original acceleration 'a' is 2 m/s², the force 'F' acting on the sled can be calculated as F=ma. But since the net force is being tripled and the mass halved, the new force would be 3F, and the new mass would be m/2.

Now we need to determine the new acceleration, which can be calculated as a = F/m. Plugging in our new values, we get a = (3F)/(m/2). This equates to a = 6F/m. Since the original force is calculated as F=ma, we can substitute this in the above equation. Hence the new acceleration is 6*2 m/s² = 12 m/s².

Learn more about Newton's second law of motion here:

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If the velocity of an object changes from 65 m/s to 98 m/s during a time interval of 12 s, what is the acceleration of the object?

Answers

a=acceleration
vf=final velocity
v₀=original velocity
t=time period

a=(vf-v₀)/t
a=(98 m/s-65 m/s)/12s=33 m/s / 12 s=2.75 m/s²

Answer: the acceleration of the object is 2.75 m/s²

     Size of acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

                             =     (98 m/s)    -    (65 m/s)  =  33 m/s .

Time for the change = 12 s

Size of acceleration =  (33 m/s) / (12 s)

                                   =    (33/12) (m/s²)

                                   =        2.75 m/s² .

The Greeks rejected the notion that the Earth orbits the Sun. Why? A.) They believed that the Sun is a God.B.) They weren't as smart as we are.
C.) They could not measure a change in stars' positions on the sky.
D.) They could not measure how big the Earth was.

Answers

B.) It ran contrary to their senses. Greeks knew that we should see stellar parallax if we orbited the Sun – but they could not detect it.

While John is traveling along an interstate highway, he notices a 194-mile marker as he passes through town. Later John passes another mile marker, 104. (a) What is the distance between town and John's current location? (b) What is John's current position?

Answers

Answer:

a) 90 miles is the distance between town and John's current location.

b) John's current position is the 104 mile marker.

Explanation:

See in the picture.

Assignment: Ultrasound and Infrasound Research Exploration Ultrasound and infrasound are categories of sound with different frequencies. These categories of sound can be used for a variety of different applications. In this research assignment, you will take a look at the industrial applications of ultrasound and infrasound. Your essay should do the following things: • Define and describe in detail (and in your own words) ultrasound and infrasound • Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples. When conducting your research, remember to gather information from multiple sources. Start your research by defining ultrasound and infrasound. Then find ideas about industrial applications. The essay should be about 350-450 words, which is 1-1.5 typed pages, double-spaced, using 12-pt. font. Check the rubric to review how you will be graded for this assignment. plz no funny answers

Answers

Answer:

Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz.  Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

Explanation:

idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest

Answer:

Ultrasound vs. Infrasound Research Exploration

Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)

-Sav xx