PHYSICS COLLEGE

true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields

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Answer 1
Answer:

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field

Related Questions

The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.
determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.
In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.
g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answers

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

K_(max)=(hc)/(\lambda)-W

Here h is the Planck's constant, c is the speed of light, \lambda is the wavelength of the light and W the work function of the element:

W=(hc)/(\lambda)-K_(max)\nW=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\nW=2.01eV

Now, we calculate the wavelength for the new maximum kinetic energy:

W+K_(max)=(hc)/(\lambda)\n\lambda=(hc)/(W+K_(max))\n\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\n\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0. Express your answer in terms of α and x.

Answers

Answer

\Delta U= \alpha (x^3)/(3) \n

Explanation:

given

F = -\alpha x^2 i  

where \alpha = 12 N/m^2

now we know

\int\limits^W_0 {} \, dW = \int\limits^a_b {F.} \, dxi ..................(i)

where dx is infinitesimal distance

W = \int\limits^a_b {-\alpha x^2} \, dx \n  

for x = a and b = 0

after integration we get

W = -\alpha (x^3)/(3)  

we know work done by conservative force will be equals to negative of potential energy

W = -\Delta U

so we get

-\Delta U= -\alpha (x^3)/(3) \n\n\Delta U= \alpha (x^3)/(3) \n

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_(initial) + PE_(initial) = KE_(final) + PE_(final)

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = (1)/(2)mv^(2) + 0

2gH = v^(2)

v = √(2* 9.8* 5) = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0* 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = (1)/(2)gt^(2)

t = \sqrt{(2h)/(g)}

t = \sqrt{(2* 2)/(9.8) = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56* 0.638 = 4.823\ m

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

mgh= (1)/(2)mv^2

v = √(2gh)

v = √(2* 9.8 * 5)

v = 9.89 m/s

now he perform elastic collision

v = (m_1v_1)/(m_1+m_2)

v = (65* 9.89)/(65+20)

v = 7.57 m/s

time taken by the skies to fall is

h = (1)/(2)gt^2

t = \sqrt{(2h)/(g)}

t = \sqrt{(2* 2)/(9.8)}

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

Answers

acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance.  If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE.  The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity.   We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball.  It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.
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What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)
Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.C) The distance between the minima stays the same.D) The distance between the minima increases.E) The distance between the maxima increases.