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4 years ago

15

Dropping a temporary magnet on the floor may destroy its magnetism because _____. the floor absorbs the magnetic field the domai

ns are knocked out of alignment the electrons start spinning the other way the heat generated by the collision destroys the magnetism

1 answer:

Svetradugi [14.3K]4 years ago

7 0

Answer: Dropping a temporary magnet on the floor may destroy its magnetism because <u>the domains are knocked out of alignment</u>

Explanation:

An object behaves as a magnet when all the domains are aligned in a single direction. A temporary magnet is created when domains are temporarily aligned in the presence of another magnetic field. When this magnet is dropped on the floor, the domains dis-align causing the demagnetization of the magnet.

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a. What is the peak wavelength for AM0? What temperature T corresponds to this peak wavelength for a blackbody source? Assuming

atroni [7]

Answer:

A) T = 5510 K , B) I = 5,226 10⁷ W / m² , C) I₂ = 1128 W / m²

Explanation:

A) The mass of air is defined as the ratio between the shortest path that sunlight must pass to reach the planet's surface and the length of the beam, is called AM

Am = 1 / cos θ

The AM0 value corresponds to solar radiation in the outer part of the Earth's atmosphere.

The peak of this emission is the peak that emitted from the sun

λ = 526 nm

To find the temperature that corresponds to this emission we use the Wien displacement law

λ T = 2,898 10⁻³

T = 2,898 10⁻³ / 526 10⁻⁹

T = 5510 K

i) The radiance on the surface of the sun is

I = P / A

We can calculate the potency by Stefan's law, for a black body

P = σ A e T⁴

P / A = σ e T⁴

The σ constant is value 5,670 10⁻⁸ W / m²K⁴, we will assume that the Sun emits as a black body, so e = 1

I = sig T⁴

I = 5,670 10⁻⁸ 5510⁴

I = 5,226 10⁷ W / m²

ii) the irradiation at a distance of 1 ua (1,496 1011 m)

Let's use the relationship

P = I A

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π R²

Let's replace

I₂ = I₁ (r₁ / r₂)²

Index 1 corresponds to the sun and the index to Earth that is an astronomical unit

r₁ = 6.96 10⁸m (Sun radius)

r₂ = 1,498 1011 m (Earth-Sun distance)

Calculous

I₂ = 5,226 10⁷ (6.96 10⁸ / 1,498 10¹¹)²

I₂ = 1.1281 10³ W / m²

I₂ = 1128 W / m²

8 0

3 years ago

If the pressure of a substance is increased during a boiling process, will the temperature also increase, or will it remain cons

7nadin3 [17]

Answer:

on increasing pressure, temperature will also increase.

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.

P ∝ T

Also,

Also, using Gay-Lussac's law,

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

Thus, on increasing pressure, temperature will also increase.

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3 years ago

The relationship between linear velocity and angular velocity

bonufazy [111]

In linear motion , when a body moves with uniform velocity , in time t , its linear displacement will be ;

S = r∅ S = vt

r∅ = vt

r.∅ / t = v

As

v = rw

where ∅ = 90° is the angle between between radius vector r and angular velocity w (omega )

In case ∅ ≠ 90° , we can write v = r w sin∅

It gives us v = w× r

7 0

3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

A)

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

B)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

D)

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F)

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

G)

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

3 years ago