I NEED HELP ASAP!!!
which of the following causes the magnetic force between the magnet and the scrap metal?

Answers

Answer 1

Answer:

Alternating current at which when entered into the loop cause it to magnetize


Related Questions

1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.

Answers

Answer:no

Explanation:because 0.9*(30*60)=0.9*1800=1620

The turtle has already won the race

Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

What will be the speed of the rabbit and the turtle?

It is given

[tex]V_{t} = 0.9 \frac{m}{s}[/tex]

[tex]V_{r} = 9 \frac{m}{s}[/tex]

[tex]D=1500 m[/tex]

Time taken by turtle  

 [tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]

[tex]T=1666 minutes= 27 hours[/tex]

Time taken by  rabbit

[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]

[tex]T=166 minutes[/tex]

since rabbit started 30 minutes after turtle then

[tex]T= 136+30=196 minutes[/tex]

[tex]T= 3.2 hours[/tex]

Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

To know more about average velocity follow

https://brainly.com/question/6504879

Why must scientists be careful when studying
nanotechnology?

Answers

Answer:

When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.

Hope it helps u:)

What is the name of the invisible line that runs
down the center of the axial region?

Answers

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.



Required information
You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN

Answers

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;

[tex]F_{min}[/tex] = m·g - m·a

∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N

A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles

Answers

Answer:

The required fraction is 0.023.

Explanation:

Given that

Mass of a car, m = 1030 kg

Mass of 4 wheels = 12 kg

We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.

The rotational kinetic energy due to four wheel is

[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]

Linear kinetic Energy of the car is:

[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]

Fraction,

[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]

So, the required fraction is 0.023.

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant

Answers

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]

A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature

Answers

Satellite video of the show will come to

A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?

Answers

Answer:

3.55 T

Explanation:

Applying,

F = BILsin∅.............. Equation 1

Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)

Substitute these values into equation 2

B = 2.45/(0.03×23×sin90)

B = 2.45/0.69

B = 3.55 T

A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?

Answers

Answer:

28 j

Explanation:

because when you add you get 28

The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.

Answers

Answer:

  f = 276.6 Hz

Explanation:

This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.

In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is

          L = λ/ 4

speed is related to wavelength and frequency

          v = λ f

          λ = v / f

we substitute

          L = v / 4f

          f = v / 4L

the speed of sound at 20ºC is

          v = 343 m / s

let's calculate

          f = [tex]\frac{343 }{4 \ 0.31}[/tex]

          f = 276.6 Hz

1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.

Answers

Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

 

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

At what angle torque is half of the max

Answers

At what angle torque is half of max

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown

Answers

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]

Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution

Answers

Answer:

the instantaneous velocity is 51 m/s

Explanation:

Given;

acceleration, a = 2 + 5t²

Acceleration is the change in velocity with time.

[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]

Therefore, the instantaneous velocity is 51 m/s

A pitching machine is programmed to pitch baseballs horizontally at a speed of 126 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. For each of the object pairings listed below, determine the correct relative speed.

a. The speed of the pitching machine relative to the truck
b. The speed of the pitched bell relative to the truck
c. The speed of the pitching machine relative to you
d. The speed of the pitched ball relative to you

Answers

Explanation:

a) zero, since the machine is mounted on the truck

b) 126 km/hr

c) 85 km/hr

d) 126 km/hr + 85 km/hr = 211 km/hr

An alternating voltage is connected in series with a resistance R and an inductance L If the potential drop across the
resistance is 200 V and across the inductance is 100V
then the applied voltage is
V 223.6
V 2006
V 300
V50
Please help me

Answers

Answer:

oh my God I got really confused right now

Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward

Answers

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B=+5\ m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]

So, the magnitude is 20 m/s and the direction is westward (negative sign).

Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm

Answers

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.

Answers

Answer:

[tex]F_b= 0.720 N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=600N[/tex]

Average density [tex]\rho=1.20kg/m^3[/tex]

Mass

[tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{600}{9.81}[/tex]

[tex]m=61.22kg[/tex]

Generally the equation for Volume is mathematically given by

[tex]V =\frac{ mass}{density}[/tex]

[tex]V= \frac{61.22}{1000}[/tex]

[tex]V=0.06122 m^3[/tex]

Therefore

Buoyant force [tex]F_b[/tex]

[tex]F_b=\rho*V*g[/tex]

[tex]F_b= rho_air*V*g[/tex]

[tex]F_b= 0.720 N[/tex]

Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction​

Answers

Answer:

a) Light that passes through the floor to reveal yourself (not shadow).

b) 2 rays of light that bounce between 2 transparent media.

c) I don't know what is Diffraction?

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing

Answers

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you

Answers

Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

find the exit angle relative to the horizontal in an isosceles triangle with 36 °​

Answers

what

what

what

what

sorry

sorrry

sorry

What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?

Answers

Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

[tex]F =\frac{mv^2}{r}[/tex]

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.


52 cm4


72 cm4


32 cm4


24 cm4


2 cm4

Answers

Answer:

Minimum Area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum Area of rectangle

Computation:

Minimum Area of rectangle = Length of rectangle x Width of rectangle

Minimum Area of rectangle = 6 x 4

Minimum Area of rectangle = 24 centimeter²

derive expression for pressure exerted by gas ​

Answers

From kinetic theory of gases, the pressure exerted by a gas is given by velocity of gas molecules. m = Mass of each molecule of a gas. But by assumptions of the kinetic theory of gases the average kinetic energy of a molecule is constant at a constant temperature.

Một ống dây điện thẳng dài có lõi sắt, tiết diện ngang của ống S = 20 cm2

, chiều dài
1 m, hệ số tự cảm L = 0,44 H. Cường độ từ trường trong ống dây là H = 0,8.103 A/m. Từ
thông gửi qua tiết diện ngang của ống bằng

3

0
1,6.10 Wb

. Cường độ dòng điện chạy

qua ống dây là

Answers

Answer:

sgsbssbduebubbeeifirjeirneejrbb8m!keoejr

d

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What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m

Answers

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is density of water

h is height

So, put all the values,

[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)

A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.

At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.

Answers

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B

Answers

Answer:

 I = - m 16  the two impulses are the same,

Explanation:

The impulse is given by the relationship

         I = Δp

         I = p_f - p₀

in this case the final velocity is zero therefore p_f = 0

        I = -p₀

For driver A the steering wheel impulse is

        I = - m v₀

        I = - m 16

For driver B, the airbag gives an impulse

        I = - m 16

We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less

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