If you drew magnetic field lines for this bar magnet, which statement would be true

Answers

Answer 1

Answer: Arrows point away from north and toward south.

and

Field lines loop around the magnet starting at the north pole and ending at the south pole.

Answer 2

Answer: field lines correct for future weiins

Related Questions

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?
Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off
Hotel rooms in Smalltown go for $100, and 1,000 rooms are rented on atypical day.a. To raise revenue, the mayor decides to charge hotels a tax of $10 perrented room. After the tax is imposed, the going rate for hotel roomsrises to $108, and the number of rooms rented falls to 900. Calculate theamount of revenue this tax raises for Smalltown and the deadweightloss of the tax. (Hint: The area of a triangle is l/2Xbase X height.)h. The mayor now doubles the tax to $20. The price rises to $116, andthe number of rooms rented falls to 800. Calculate tax revenue anddeadweight loss with this larger tax. Are they double, more thandouble, or less than double? Explain.
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?

Answers

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

$\beta=(\lambda D)/(d)$

Put the value into the formula

$\beta=(0.25*10^(-9)*3.3)/(0.16*10^(-3))$

$\beta=5.15*10^(-6)\ m$

$\beta=5.15\ \mu m$

Hence, The spacing is 5.15 μm.

Final answer:

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

Explanation:

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.

Learn more about Double Slit Interference here:

brainly.com/question/32574386

#SPJ11

The ____ contains the highest concentration of ozone

Answers

I believe the term you are looking for is the ozone layer. This layer is in the highest region if the stratosphere.

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Part a)

Equation of position with time is given as

$x = (2.0 m/s)t + (-3.0 m/s2)t^2$

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

$x = 2* 0.45 - 3 * 0.45^2$

$x_1 = 0.2925 m$

at t = 0.55 s

$x = 2* 0.55 - 3*0.55^2$

$x_2 = 0.1925$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.1925 - 0.2925 = -0.1 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.1}{0.1) = -1 m/s$

part c)

at t = 0.49 s

$x = 2* 0.49 - 3 * 0.49^2$

$x_1 = 0.2597 m$

at t = 0.51 s

$x = 2* 0.51 - 3*0.51^2$

$x_2 = 0.2397 m$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.2397 - 0.2597 = -0.02 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.02}{0.02) = -1 m/s$

A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min

Answers

Answer:

6.4 rpm

Explanation:

$I_(m)$ = moment of inertia of merry-go-round = 275 kgm²

m = mass of the child = 23 kg

R = radius of the merry-go-round = 2.20 m

$I_(c)$ = moment of inertia of child after jumping on merry-go-round = mR² = (23) (2.20)² = 111.32 kgm²

Total moment of inertia after child jumps is given as

$I_(f)$ = $I_(m)$ + $I_(c)$ = 275 + 111.32 = 386.32 kgm²

Total moment of inertia before child jumps is given as

$I_(i)$ = $I_(m)$ = 275 kgm²

$w_(i)$ = initial angular speed = 9 rpm

$w_(f)$ = final angular speed

using conservation of angular momentum

$I_(i)$ $w_(i)$ = $I_(f)$ $w_(f)$

(275) (9) = (386.32) $w_(f)$

$w_(f)$ = 6.4 rpm

Maggots feed on dead and decaying organisms for energy. What are maggots?autotrophs
producers
decomposers
heterotrophs

Answers

Answer:

Explanation:

Decomposers is the correct answer

Answer:

Decomposers is the right answer

Explanation:

Maggots are decomposers because they eat the dead bodys for energy

I don't know if the thing I wrote it truse so ya

Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?
b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers

Answers

Answer:

a) It will be easier to apply all her force as far as possible from the axle

b) There is a net torque applied

Explanation:

a) Applying a larger force far away from the axle of the door produces a larger force on the torque than pushing it near the axle. This is because Increasing the lever arm between the axle and the point at which you push the door increases the torque on the door

b)The door is turning faster and faster because you are exerting a torque on it and its undergoing angular acceleration. There is a net torque which is the addition of the torque applied on the opposite sides of the door

Other Questions

22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)2N250N5000N50000N

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?A. FB. 4FC. 4F/3D. 4F/9E. F/3

A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?

Does lighting striking the earth considered the speed of light?