The primary coil of a transformer is connected to a 120 v wall outlet. the secondary coil is connected to a lamp that dissipates 60 w. what is the nns current in the primary coil?

Answers

Answer 1

Energy/power is not gained or lost going through a (ideal) transformer.

So the transformer in this problem really doesn't matter.  If the lamp is using energy at the rate of 60 watts, then the whole contraption is getting 60 watts of power from the wall outlet.

Power = (voltage) x (current)

60 watts = (120 v) x (current)

Current = (60 watts) / (120 v)

Current = 0.5 Ampere


Related Questions

Susie rubbed the plastic wand with a silk cloth. She brought the wand close to the top of the electroscope. Note what happened when she did this. How did Susie make the initial static charge that started this process?

Answers

Answer:

She caused friction

Explanation:

(I'm not a 100% sure but, whatever) When you rub something up against another object, it causes friction. An example is rubbing a balloon to your hair. It sticks to your hair because it has (I think) like charges. Hope this helped

Answer:

She got initial negative charge on wand due to friction.

Explanation:

When Susie rubbed silk with the plastic wand, she makes static negative charge on the plastic wand due to friction. When she brought this negatively charged wand near the electroscope, electrons are pushed down into the electroscope. As a result, conducting rod and foil of electroscope become negatively charged but, net charge on electroscope is still zero.  

A car has a mass of 1300 kg and a velocity of 15 m/s. The car crashes into a wall and stops in two seconds. What is the stopping force during the collision?
A: 4875 N
B: 9750 N
C: 19500 N
D: 39000 N

Answers

Answer:

B

Explanation:

Newton's second law:

F = ma

The acceleration is Δv/Δt, so:

F = m Δv/Δt

Given m = 1300 kg, Δv = 15 m/s, and Δt = 2 s:

F = (1300 kg) (15 m/s) / (2 s)

F = 9750 N

Answer B.

How many neutrons are needed to initiate the fission reaction shown?

Answers

Answer: One neutron

Explanation:

one neutron 1/0n

Sum up the mass numbers on the right 99 + 135 + 2 = 236.

The sum of the mass numbers on the left should equal 236. 235 + 1 = 236

Final answer:

Only one neutron is needed to initiate a fission reaction in Uranium-235. This initiates a chain reaction where released neutrons cause further fission. However, not all neutrons result in further fission, as some may escape or interact without causing a split.

Explanation:

To initiate a fission reaction in Uranium-235, only one neutron is needed. As the process begins, the uranium-235 nucleus absorbs a neutron, making it an unstable uranium-236 nucleus. This unstable nucleus then breaks down into two smaller nuclei, releasing a large amount of energy in the process. Additionally, two or three neutrons are also released during the breakdown of the unstable uranium-236 nucleus. These extra neutrons can in turn go on to initiate fission in other uranium-235 nuclei, leading to a nuclear chain reaction. However, not every neutron produced results in further fission as some neutrons may escape the material or interact with a nucleus without causing it to split.

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Mariner 10 was the first to visit this planet in 1974. Which planet is this?

Answers

Answer: Mercury

The Mariner 10 probe was launched by NASA on November 3rd, 1973, with the purpose of exploring the characteristics of two planets in the solar system that were closest to the Sun, Mercury and Venus.

In addition, it was launched to explore the atmosphere and surface of both planets and prove that it was possible to use gravitational assistance (also called slingshot effect, a special orbital maneuver in order to use the gravitational field energy of a planet or massive body to accelerate or slow the probe and change the direction of its trajectory) in long interplanetary trips to save fuel.

In this case, Mariner 10 first arrived at Venus and succeded in using its gravitational field to accelerate its trajectory towards Mercury.

Mariner 10 was the first spacecraft to visit Mercury in 1974, transmitting over 2000 detailed photographs of the planet's surface.

The planet that Mariner 10 first visited in 1974 is Mercury. On its flyby, Mariner 10 passed 9,500 kilometers from the surface of Mercury and sent back more than 2000 photographs. These photographs were groundbreaking, offering detailed views with a resolution down to 150 meters and marking a milestone in space exploration. Mercury is the 1st planet in the solar system of Milky way and it is comparatively much smaller than the Earth.

The distance traveled by an object divided by the time it takes to travel that distance is called

Answers

Answer:

Speed

Explanation:

- Speed is a scalar quantity that represents the rate of change of distance. It is calculated as

[tex]v=\frac{d}{t}[/tex]

where

d is the distance travelled by the object (regardless of its direction)

t is the time elapsed

The speed is measured in meters per second (m/s). We can also notice that speed is different from velocity: in fact, speed is a scalar quantity (magnitude only), while velocity is a vector quantity (magnitude+direction).

Many sophisticated cameras have zoom lenses. When you select the telephoto setting, the objects in front of you appear much closer. In this setting, (A) the focal length of the lens is shorter than normal. (B) the aperture of the lens is larger than normal. (C) the focal length of the lens is longer than normal. (D) the aperture of the lens is smaller than normal.

Answers

The amswer is a i think cuz it make semse

Which is the hottest planet in the solar system

Answers

Answer:

venus

Explanation:

         Answer:

► Venus

         Explanation:

Venus is the hottest planet in the Solar System. It is not the closest, but it is the hottest. Venus's temperature has an average if 462 degrees Celsius. That is 863.6 Fahrenheit.

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center?a) 7/5 Ib) 3/5 Ic) 2/5 I
d) 1/7 I

Answers

Answer:

2/7 I

Explanation:

The theorem of parallel axis states that the moment of inertia of a body about a certain axis z' is equal to the moment of inertia of the body about the axis passing through the centre, z, plus the product between the mass of the body (M) and the square of the distance (r) between the two axis:

[tex]I_z' = I_z + Mr^2[/tex] (1)

For a solid sphere, the moment of inertia about the axis passing through the centre is

[tex]I_z=\frac{2}{5}MR^2[/tex] (2)

where R is the radius of the sphere.

The moment of inertia about an axis tangent to the surface then will be (applying (1) using r=R):

[tex]I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2[/tex] (3)

The problem asks us to rewrite [tex]I_z[/tex], the moment of inertia about the centre, in terms of I, the moment of inertia about the axis tangent to the surface. We can do it by rewriting (2) as follows:

[tex]MR^2 = \frac{5}{2}I_z[/tex]

And substituting this into (3):

[tex]I=\frac{7}{5}(MR^2 )=\frac{7}{5}(\frac{5}{2} I_z) = \frac{7}{2}I_z\\I_z = \frac{2}{7}I[/tex]

Final answer:

The moment of inertia of a uniform solid sphere about an axis through its center is 3/5 times the moment of inertia of the sphere about an axis tangent to its surface. This is calculated using the parallel axis theorem.

The correct option is b.

Explanation:

In physics, the moment of inertia of a sphere about an axis through its center is determined using the parallel axis theorem. The formula of the parallel axis theorem is: Icm = I + Mh2, where Icm is the moment of inertia about an axis through the center of mass, I is the moment of inertia about a parallel axis through the edge of the sphere, M is the mass of the sphere, and h is the distance between the two axes.

In this case, the sphere is uniform, so its center of mass is in its geometric center. The axis through the edge of the sphere is a distance of the radius of the sphere away from the axis through its center, so h = r. Also, in a solid sphere, the moment of inertia, Icm, about an axis through its center is (2/5)MR2.

With these values substituted into the formula, we have: (2/5)MR2 = I + MR2.

From this it can be deduced that I = 2/5 MR2 - MR2 = -(3/5) MR2. So the moment of inertia of this sphere about an axis through its center is (3/5) times smaller than the moment of inertia about an axis tangent to its surface.

This gives us an answer of choice b) 3/5 I.

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The temperature of a chemical reaction ranges between −10 degrees Celsius and 50 degrees Celsius. The temperature is at its lowest point when t = 0, and the reaction completes 1 cycle during a 6-hour period. What is a cosine function that models this reaction?

Answers

Answer:

Explanation:

The general cosine wave function is:

y = A cos(ωt + φ) + B

where A is the amplitude, ω is the frequency, φ is the phase offset, and B is the vertical offset.

The temperature ranges from -10 to 50, so the magnitude of the amplitude is:

|A| = (50 − -10) / 2

|A| = 30

It's at its lowest point at t=0, so the sign of the amplitude is -1:

A = -30

The lowest point, -10, is 20 more than the amplitude, so the vertical offset is:

B = 20

The reaction completes 1 cycle in 6 hours, so f = 1/6.  ω = 2πf, so:

ω = 2π (1/6)

ω = π/3

So the function is:

y = -30 cos(π/3 t) + 20

Alternatively, instead of a negative sign, we could have added a phase shift of π:

y = 30 cos(π/3 t + π) + 20

Either of these answers is correct.

Planet with an atmosphere that rains sulfuric acid

Answers

Answer:

Venus

Explanation:

The mass of a proton is approximately equal to

Answers

Answer:

[tex]1.6726\cdot 10^{-27} kg[/tex]

Explanation:

The three main particles that make an atom are:

- Proton: its mass is [tex]1.6726\cdot 10^{-27} kg[/tex], it carries an electric charge of +e ([tex]e=1.6\cdot 10^{-19}C[/tex]), and it is located in the nucles of the atom

- Neutron: its mass is [tex]1.6749 \cdot 10^{-27}kg[/tex], it carries no electric charge, and it is also located in the nucleus of the atom

- Electron: its mass is [tex]9.1094 \cdot 10^{-31}kg[/tex], it carries an electric charge of -e ([tex]e=1.6\cdot 10^{-19}C[/tex]), and it is located outside the nucleus

Why is the nervous system like a telegraph

Answers

Answer: The structure of axon bundle in nervous system is just like a telegraph wire.

Explanation: There is structural and functional similarity between the nervous system and telegraph. In a telegraph the wire of the cable are bundled to form a single cable just like the way a group of axons bundle themselves.

Also the axons are covered by a myelin (also known as white matter) to insulate them in a similar way as the plastic coating of an electric wire of telegraph.  Both axon in the nervous system and the telegraph wire send signal across its ends.  

What type of simple machine is a catapult

Answers

Answer: A lever

The catapult is basically a type of simple lever or first class lever, which is a device used to transmit force and displacement to an object, by means of the amplification of the applied mechanical force, thus increasing its speed or distance traveled.

These simple levers are composed of a rigid bar that can rotate freely around a point of support (the fulcrum). However what differentiates them from the other levers is that the fulcrum is between the point where the effort must be applied and the point where the resistance  is.

With this configuration it is posssible to make several arrangements, depending on the purpose to be achieved, either control and decrease the speed and distance traveled by the object or increase it.

A catapult is a simple machine that functions largely as a lever, one of the classical types of simple machines that provides a mechanical advantage by altering force and distance.

A catapult is a simple machine that utilizes the principles of a lever to multiply the force applied to it. Simple machines are devices that can be used to multiply or augment a force that we apply. The catapult lever operates by converting stored energy into kinetic energy, effectively using the conservation of energy principle. When designing a  small catapult to try at home, you can use materials such as rubber bands, spoons, popsicle sticks, or small plastic containers to create a device that demonstrates this principle.

Simple machines like the lever, nail puller, wheelbarrow, and crank are designed to give us a mechanical advantage. This involves changing the magnitude or direction of forces, helping to perform tasks more easily by requiring less force over a greater distance. Catapults, in particular, allow a small input force to be converted into a much larger output force, launching projectiles over a distance.

What occurs during a solar eclipse? Check all that apply.


Earth is closest to the Sun.


There are small tides across Earth.


The moon’s shadow falls on Earth.


The moon is covered in Earth’s shadow.


The moon is between Earth and the Sun.

Answers

Answer:

The Moon’s shadow falls on Earth.

The Moon is between Earth and the Sun.

Explanation:

Eclipses are known as game of shadows. During an eclipse shadow of an object falls on another object. From the perspective of Earth we can see two kind of eclipses: Lunar and Solar.

In Solar eclipse, the Moon lies in between our planet the Earth and the Sun such that they are in the same line. Shadow of Moon will fall on Earth. Looking from the Earth, the Sun will look like it has been hidden by the Moon.

Final answer:

A solar eclipse occurs when the Moon moves between the Sun and Earth, casting a shadow on Earth. Total solar eclipses happen when the Moon's umbra reaches Earth, while partial eclipses occur within the penumbral shadow. Lunar eclipses differ as they involve the Moon entering Earth's shadow.

Explanation:

During a solar eclipse, the Moon moves between the Sun and Earth, casting its shadow on our planet. If the eclipse is total, it occurs when the umbra, the Moon's darkest shadow, reaches the surface of the Earth, covering the Sun completely for a brief time. This results in the solar atmosphere, known as the corona, becoming visible. During the eclipse, the observers within the penumbra, a lighter shadow, will see only a partial covering of the Sun, known as a partial solar eclipse. A solar eclipse requires a specific alignment on the ecliptic plane, which is the plane of Earth's orbit around the Sun. In contrast, a lunar eclipse takes place when the Moon passes into Earth's shadow and is visible from the entire night hemisphere of our planet.

Regarding the options given in the question:

The Moon's shadow falls on Earth. TrueThe Moon is between Earth and the Sun. True

The notions that Earth is closest to the Sun and that the Moon is covered in Earth's shadow describe other phenomena, not a solar eclipse. Additionally, tides may be affected during a solar eclipse but are not specifically mentioned in the context of the eclipse itself.

A constant eastward horizontal force of 70. newtons is applied to a 20.-kilogram crate moving toward the east on a level floor. If the frictional force on the crate has a magnitude of 10. newtons, what is the magnitude of the crate’s acceleration?

Answers

Answer:

[tex]3 m/s^2[/tex]

Explanation:

Newton's second law states that:

[tex]\sum F = ma[/tex]

where

[tex]\sum F[/tex] is the resultant of the forces acting on an object

m is the mass of the object

a is the acceleration

In this case we have two forces:

F = +70 N (east direction) is the horizontal push

Ff = -10 N (west direction) is the frictional force

so the net force is

[tex]\sum F=70 N - 10 N = 60 N[/tex]

We also know the mass of the crate

m = 20 kg

So we can find the acceleration

[tex]a=\frac{sum F}{m}=\frac{60 N}{20 kg}=3 m/s^2[/tex]

The energy absorbed or released during a reaction in which a substance is produced is called the

Answers

the answer is bond energy but I am not pretty sure

Which of these is a correct equation for acceleration? A. a = m × F B. a = m + F C. a = F ÷ m D. a = m ÷ F

Answers

D. Force is equal to mass times acceleration.

F=ma

m/F=a

a = F ÷ m is the correct equation for acceleration. Hence, option (A) is correct.

What is acceleration?

Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.

Acceleration is a vector quantity since it has both a magnitude and a direction. A vector quantity is also velocity. The velocity vector change during a time interval divided by the time interval is the definition of acceleration.

According to Newton's second law of motion:

Force = mass × acceleration

Hence, acceleration (a) = force (F) ÷ mass (m).

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_______________Is the distance traveled during a specific unit of time.

Answers

Answer:

Speeed

Explanation:

The speed of an object is defined as the rate of change of distance:

[tex]v=\frac{d}{t}[/tex]

where

d is the distance covered by the object

t is the amount of time needed to cover that distance

Speed is measured in meters per second (m/s). It should be noted that speed is a scalar quantity, so it only has a magnitude (and no direction).

When a solid is hit hard, the force may break the solid. This is called

Toughness

Cleavage

Hardness

Answers

Answer:

Cleavage

Explanation:

Toughness is the resistance of a solid to breakage. The tougher a solid is, the less likely it would succumb to breakage of any form.

Hardness is very similar to toughness and it describes how unyielding a sold is to pressure or force of any form.

Cleavage on the other hand describes the tendency or ability of a solid to spilt or be broken along a specified plane of weakness. When such a solid is hit hard, they show some fracturing directions which represents weaknesses in their lattice structures or crystal forms. Most solids have cleavage directions in them.

Pls help on this one

Answers

Answer:

D

Explanation:

because 50.0/10.0 = 5.0

Answer:

D

Explanation:

because 50.0/10.0 = 5.0

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T.

a.) If the bob's mass is doubled, approximately what will the pendulum's new period be?
b.) If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now be?
c.) If the pendulum is taken into the orbiting space station what will happen to the bob?

Answers

The answer to this question is b

The right question is

b.) If the pendulum is carried to the moon where the acceleration of gravity is around g / 6, what is the current period?

Further Explanation

A simple pendulum consists of a light string and a small ball (pendulum ball) with mass m hanging from the end of the rope. In analyzing the movement of a simple pendulum, the air friction force is ignored and the mass of the rope is so small that it can be ignored relative to the ball.

A simple pendulum consisting of a rope with a length L and a pendulum ball with mass m. The forces acting on the pendulum ball are the weight force (w = mg) and the FT string tension force. Gravity has a component of mg cos theta which is in the direction of the rope and mg sin theta which is perpendicular to the rope. The pendulum oscillates due to the presence of mg sin theta gravity component. Because there is no air friction, the pendulum oscillates along a circular arc with the same amplitude.

The requirement for an object to do Simple Harmonic Motion is if the recovery force is proportional to the deviation. If the recovery force is proportional to the deviation of x or the angle of the theta, the pendulum performs Simple Harmonic Motion.

The simple pendulum period can be determined using the equation:

T = 2n (sqrt m / k

We replace the effective force constant with mg / L

T = 2n (sqrt m / (mg / L))

T = 2n (sqrt L / g -> 0 small)

Simple Pendulum Frequency

f = 1 / T

f = 1 / 2n (sqrt L / g)

f = (1 / 2n) (sqrt g / L -> 0 small)

This is a simple pendulum frequency equation

Information :

T is the period, f is the frequency, L is the length of the rope and g is the acceleration due to gravity.  

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Class: High School

Subject: Physics

Keywords: pendulum, simple, formula

The equation h=80t-16t^2 models the height h in feet reached in t seconds by an object propelled straight up from the ground at a speed of 80 ft/s. use the discriminant to find whether the object will ever reach a height of 90 ft

Answers

The object will reach a height of 90ft

To solve this exercise we are going to use  the discriminant of the quadratic polynomial ax²+bx+c=0, which is b²-4ac.  

If the discriminant is negative, then there are no real solutions to the equation.

If the discriminant is zero, there is only one solution.

If the discriminant is positive, there are two real solutions.

We have the equation h(t)=18t-16t² which describes the model of the height h (feet) reached in t (seconds) by an object propelled straight up from the ground at a speed of 80 ft/s. We want to use the discriminant to find whether the object will ever reach a height of 90ft.

First, we have to rewrite the equation to the form ax²+bx+c=0 and we know the height that is possible to reach for the object h=90ft.

90 = 80t-16t² ---------->  -16t²+80t-90=0

Using the discriminan equation D = b²- 4ac.

From the quadratic polynomial -16t²+80t-90=0, we have a = -16, b = 80, and c = -90

D = (80)² - 4 (-16)(-90)

D = 6400 - 5760 = 640

Since the discriminant D is positive, the object will reach a height of 90ft.

A cart is attached to a string and the other end of the string is attached to a weight that is free to fall. The cart is initially released from rest, and travels a distance of 0.521 m m before hitting a stopper. The cart's final speed was 0.931 m/s . The inertia of the weight is 0.115 kg , while the inertia of the cart is 1.000 kg . For the next 4 parts consider a system of the earth, the weight, the cart, the string, and the track. Assume the track does not move relative to the earth.1- What is the final speed of the weight?2- What is the change in kinetic energy of the system?3- What is the change in potential energy of the system?4- What is the change in thermal energy of the system due to friction?

Answers

1. 3.20 m/s

Assuming the string is inextensible, the cart and the weight travels the same distance: so, since the cart travels for 0.521 m, the distance travelled by the weigth is the same:

d = 0.521 m

The motion of the weigth is a free-fall motion with acceleration g = 9.8 m/s^2, so its final speed can be found by using the equation

[tex]v^2 = u^2 + 2gd[/tex]

where

u = 0

is the initial speed of the weigth (at rest). Substituting into the formula, we find

[tex]v=\sqrt{0+2(9.8 m/s^2)(0.521 m)}=3.20 m/s[/tex]

2. 1.022 J

The change in kinetic energy of the system is equal to the sum of the kinetic energies acquired by the cart and the weight. They both started from rest, so their initial kinetic energies were zero.

The cart has

mass: m = 1.000 kg

final speed: v = 0.931 m/s

so its gained kinetic energy is

[tex]K_c = \frac{1}{2}mv^2=\frac{1}{2}(1.000 kg)(0.931 m/s)^2=0.433 J[/tex]

The weight has

mass: m = 0.115 kg

final speed: v = 3.20 m/s

so its gained kinetic energy is

[tex]K_w = \frac{1}{2}mv^2=\frac{1}{2}(0.115 kg)(3.20 m/s)^2=0.589 J[/tex]

So the change in kinetic energy of the system is

[tex]\Delta K= K_f - K_i = (0.433 J+0.589 J)-0=1.022 J[/tex]

3. -5.693 J

The potential energy of the falling cart decreases by the following amount:

[tex]\Delta U = mg \Delta h[/tex]

where

m = 1.000 kg is the mass

g = 9.8 m/s^2

[tex]\Delta h = -0.521 m[/tex] is the change in height of the cart

Substituting, we find

[tex]\Delta U=(1.000 kg)(9.8 m/s^2)(-0.521 m)=-5.106 J[/tex]

the potential energy of the falling weight decreases by the following amount:

[tex]\Delta U = mg \Delta h[/tex]

where

m = 0.115 kg is the mass

g = 9.8 m/s^2

[tex]\Delta h = -0.521 m[/tex] is the change in height of the weight (calculated at point a)

Substituting, we find

[tex]\Delta U=(0.115 kg)(9.8 m/s^2)(-0.521 m)=-0.587 J[/tex]

So, the change in potential energy is

[tex]\Delta U = -5.106 J-0.587 J=-5.693 J[/tex]

4. +4.671 J

The change in thermal energy of the system due to friction is equal to the loss in mechanical energy of the system.

The system has gained a kinetic energy equal to

[tex]\Delta K=+1.022 J[/tex]

while it has lost a potential energy equal to

[tex]\Delta U =-5.693 J[/tex]

So the loss in mechanical energy of the system is

[tex]\Delta E=\Delta K+\Delta U=+1.022 J-5.693 J=-4.671 J[/tex]

So the change in termal energy is

[tex]\Delta E_{th} = -(\Delta E)=-(-4.671 J)=+4.671 J[/tex]

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. A typical axon carries a peak current of 0.040 ?A.

What is the strength of the field at a distance of 1.6 mm ?

Answers

Answer:

[tex]5.0\cdot 10^{-12}T[/tex]

Explanation:

We can think the axons as current-carrying wires

The strength of the magnetic field produced by a current-carrying wire is

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

[tex]\mu_0[/tex] is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

[tex]I=0.040 \mu A = 0.04\cdot 10^{-6} A[/tex]

r = 1.6 mm = 0.0016 m

So the strength of the magnetic field is

[tex]B=\frac{(4\pi \cdot 10^{-7}H/m)(0.04\cdot 10^{-6} A)}{2\pi (0.0016 m)}=5.0\cdot 10^{-12}T[/tex]

Kepler's laws follow which law discovered by Sir Isaac Newton?

Answers

Kepler found that the orbits of the planets flowed three laws (rayed disk),(black dot) and maybe (symbols) hopes this helps out for you

Kepler took Brahe's detailed descriptions and measurements of the motion of the planets in the sky over many years, and derived his 3 laws of Planetary Motion based purely on what Brahe saw.

A hundred years AFTER Kepler, Newton proposed his Law of Universal Gravitation.

Using his own Law of Universal Gravitation, along with his laws of motion, Newton showed that IF his "laws" were correct, then planets HAVE TO move exactly according to Kepler's laws.  (He had to invent Calculus in order to demonstrate this.)

This was an awesome, powerful confirmation of Kepler's work and Newton's work.  Boat uvum.

In my Physics courses, I used to be able to take Newton's laws of gravity and motion, fold in some calculus and some geometry, mix until smooth, and derive Kepler's laws of planetary motion.  But that was long ago, in a galaxy far away, and, sadly, ya don't get to use it very often as an Electrical Engineer.  So I imagine it's still true, but I can't prove it now.

Three disks are spinning independently on the same axle without friction. Their respective rotational inertias and angular speeds are I,ω (clockwise); 2I,3ω (counterclockwise); and 4I,ω/2 (clockwise). The disks then slide together and stick together, forming one piece with a single angular velocity. What will be the direction and the rate of rotation ωnet of the single piece? Express your answer in terms of one or both of the variables I and ω and appropriate constants. Use a minus sign for clockwise rotation.

Answers

Answer:

3/7 ω

Explanation:

Initial momentum = final momentum

I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet

-Iω + 6Iω - 2Iω = 7I ωnet

3Iω = 7I ωnet

ωnet = 3/7 ω

The final angular velocity will be 3/7 ω counterclockwise.

Answer:

[tex]\omega_{net} = 3\omega/7[/tex]

Explanation:

For this problem we will use the conservation of angular momentum. This is, the momenta of each disk added together is equal to the momenta of the single piece at angular velocity [tex]\omega_{net}[/tex]. If

[tex]L_{0} = -I\omega-4I\frac{\omega}{2}+2I(3\omega) \\L_{0} = -I\omega-2I\omega+6I\omega \\L_{0} = 3I\omega\\[/tex],

and because all disks are spinning on the same axle, the total inertia moment of the single piece at angular velocity [tex]\omega_{net}[/tex] is the sum of the inertia moment of the three disks. This way, we have that

[tex]L_{f} = (I+2I+4I)\omega_{net}\\\\L_{f}=7I\omega_{net}\\[/tex].

The conservation of angular momentum leads us to

[tex]L_{0}=L_{f}\\[/tex],

[tex]3I\omega = 7I\omega_{net}\\[/tex],

thus

[tex]\omega_{net} = \frac{3}{7}\omega[/tex].

please help on this one?

Answers

Object distances are always negative according to the Cartesian sign convention

When did robert fulton invent the steamboat

Answers

Answer:

1807

Explanation:

Robert Fulton (1765–1815) was an American engineer and inventor who is widely known for developing a commercially successful steamboat called Clermont. In 1807, that steamboat took passengers from New York City to Albany and back again, a round trip of 300 miles, in 62 hours.

Final answer:

Robert Fulton invented the steamboat engine in 1807, significantly improving water transportation with the Clermont on the Hudson River, and facilitating economic growth and western settlement.

Explanation:

Robert Fulton invented the steamboat engine which was utilized in his first successful commercial steamboat, the Clermont, in 1807. Operating on the Hudson River, the Clermont was influential in transforming water transportation by allowing more reliable and quicker travel independent of wind. It traveled from New York City to Albany in a mere 32 hours. Fulton's innovation prompted widespread economic development, particularly in the Mississippi River Valley, and revolutionized the settlement of the West. By the 1830s, over a thousand steamboats were in operation. Yet, steamboats were prone to dangers such as boiler explosions, which eventually led to safety regulations.

A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. the electric field between the plates is increasing at the rate 1.0 * 106 v/m s. what is the magnetic field strength (a) on the axis, (b) 3.0 cm from the axis, and (c) 7.0 cm from the axis?

Answers

(a) 0

The magnetic field strength insidea a parallel-plate capacitor with changing electric field can be found by applying Ampere's law:

[tex](2\pi r) B = \mu_0 I_D[/tex] (1)

where

[tex](2\pi r)[/tex] is the circumference of the circular line of radius r with axis coincident to the axis of the capacitor, used to calculate the magnetic field

B is the strength of the magnetic field

[tex]I_D[/tex] is the displacement current enclosed by the area of the circular line mentioned above, and it is equal to

[tex]I_D = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 (\pi r^2) \frac{dE}{dt}[/tex] (2)

where

[tex]\frac{d\Phi_E}{dt}[/tex] is the rate of change of electric flux through the area enclosed by the line

[tex]\frac{dE}{dt}=1.0\cdot 10^6 V/m[/tex] is the rate of change of the electric field

Rewriting eq.(1), we find

[tex]B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}[/tex]

which is valid for r < R (where R=5.0 cm is the radius of the plates of the capacitor).

In this part of the problem,

r = 0

since we are on the axis; so substituting r=0 inside the formula above, we find

B(0) = 0

(b) [tex]1.67\cdot 10^{-13}T[/tex]

In this part, we have

r = 3.0 cm = 0.03 m

The formula used in part (a) is still valid since r<R, so we can directly use it to find the magnitude of the magnetic field:

[tex]B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.03 m)}{2}(1.0\cdot 10^6 V/m)=1.67\cdot 10^{-13}T[/tex]

(c) [tex]1.98\cdot 10^{-13} T[/tex]

In this part, we have

r = 7.0 cm = 0.07 m

so here

r > R

therefore we need to substitute [tex](\pi r^2)[/tex] with [tex](\pi R^2)[/tex] in eq. (2), since the area through which the flux is calculated is only [tex](\pi R^2)[/tex] (there is no electric field outside the area of the capacitor). So we find

[tex]I_D = \epsilon_0 (\pi R^2) \frac{dE}{dt}[/tex]

and therefore

[tex]B = \frac{\mu_0 \epsilon_0 R^2}{2r}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.05 m)^2}{2(0.07 m)}(1.0\cdot 10^6 V/m)=1.98\cdot 10^{-13} T[/tex]

The magnitude of the magnetic field 3 cm from the axis is  [tex]1.67\times 10^{-13} \rm \ T[/tex].

What is a magnetic field?

It is a vector field in which ferromagnetic objects and moving charges experience an influence.

The magnitude of the magnetic field can be calculated by the formula,

[tex]B =\dfrac { \mu _0 \epsilon_0 r} 2\times \dfrac {dE}{dt}[/tex]


Where,

μ -  magnetic permeability = [tex]4\pi \times 10^{-7}{\rm \ H/m}[/tex]

r - distance = 3 cm

[tex]\dfrac {dE}{dt}[/tex] -  rate of electic field =  [tex]1.0 \times 10^6 \rm v/m s.[/tex]

Put the values in the formula,

[tex]B =\dfrac { 4\pi \times 10^{-7}{\rm \ H/m} (8.85\times 10^{-12}) (0.03)} 2\times(1.0\times 10^6)\\\\B = 1.67\times 10^{-13} \rm \ T[/tex]

Therefore, the magnitude of the magnetic field 3 cm from the axis is  [tex]1.67\times 10^{-13} \rm \ T[/tex].

Learn more about the magnetic field:

https://brainly.com/question/19542022

The pole star, polaris, is nearby stationary and straight overhead when seen from the north pole. when viewed from the equator it

Answers

-- The 'pole star', Polaris, is always nearly stationary in the sky.  

-- It's located almost straight overhead when seen from the north pole.  

-- When viewed from the equator, Polaris is right on the northern horizon.

-- Looking from anywhere in the southern hemisphere (south of the equator), Polaris is below the horizon, and can't be seen at all.  

A person who was born and raised in the southern hemisphere, and who has never crossed the equator, has never seen Polaris, the "North Star" !

Final answer:

Polaris, known as the pole star, appears nearly overhead at the North Pole. As one moves towards the equator, Polaris drops closer to the horizon and is directly on the northern horizon when viewed from the equator. This appearance changes due to Earth's rotation and the precession of the equinoxes, meaning Polaris won't always be the pole star.

Explanation:

The pole star, Polaris, occupies a special position in the sky nearly aligned with Earth's rotational axis. At the North Pole, Polaris appears almost directly overhead. However, as one travels towards the equator, the angle at which Polaris is seen decreases. This is because the celestial sphere appears to turn around the Earth's axis, and Polaris is situated close to the north celestial pole. Thus, at the equator, Polaris is positioned right at the northern horizon, and as one goes further south,

it is no longer visible. Instead, one can observe the southern celestial pole. It is interesting to note that the close alignment of Polaris with the north celestial pole is temporary in the grand scheme of Earth's history due to the precession of the equinoxes. In the past, other stars, like Thuban, have served as the pole star, and in the future, Polaris will no longer hold that position.

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