PHYSICS COLLEGE

What is a light year

Answers

Answer 1

Answer:

Answer:

A light-year is the distance light travels in one year.

Answer 2

Answer:

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).

Related Questions

A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)
A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXABioelectrical impedanceSkinfold testingHydrostatic weighing
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answers

Answer:

Explanation:

Given

Radius of Pulley r=12 cm

mass of block m=60 gm

mass of Pulley M=430 gm

Block descend h=50 cm

Applying Conservation of Energy

Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block

i.e.

$mgh=(1)/(2)I\omega ^2+(1)/(2)mv^2$

where I=moment of inertia

$I=mr^2$

and for rolling $\omega =(v)/(r)$

$mgh=(1)/(2)Mv^2+(1)/(2)mv^2$

$v^2=(2mgh)/(m+M)$

$v=\sqrt{(2mgh)/(m+M)}$

$v=\sqrt{(2* 60* 9.8* 0.5)/(430+60)}$

$v=\sqrt{(60* 9.8)/(490)}$

$v=√(1.2)$

$v=1.095 m/s$

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

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a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

$E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J$

So, the change in its gravitational potential energy is 1.76 J.

A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m

Answers

Answer:

The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

Explanation:

Given that,

Velocity = 270 m/s

Acceleration = 9.0g s

We need to calculate the altitude

Using formula of centripetal acceleration

$a_(c)=(v^2)/(r)$

$r=(v^2)/(a_(c))$

Where, v = velocity

r = altitude

a = acceleration

Put the value into the formula

$r=(270^2)/(9.0*9.8)$

$r=826.53\ m$

Hence, The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus. What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

Answers

Answer:

a) F = -1.82 10⁻¹⁵ N, b) K = 9.1 10⁻¹⁶ J

Explanation:

a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation

F = k q Q / r²

as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual

let's calculate

F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²

F = -1.82 10⁻¹⁵ N

b) they ask us for kinetic energy

let's use Newton's second law

F = m a

acceleration is centripetal

a = v² / r

we substitute

F = m v² / r

v = √ (F r / m)

v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)

v = √ (0.2 10⁻¹⁶)

v = 0.447 10⁸ m / s

kinetic energy is

K = ½ m v²

K = ½ 9.1 10⁻³¹ (0.447 10⁸)²

K = 0.91 10⁻¹⁵ J

K = 9.1 10⁻¹⁶ J

If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.

Answers

Answer:

0.25( m1m1) , 0.75( m2m2)

Explanation:

Noted the formula for kinetic energy is 1/2(Mass × Velocity).

Therefore the original value of the mass is 0.5, giving half away makes it 0.25 to another mass which is primarily 0.5. This now makes the new mass 0.25+0.5=0.75.

Thank you.

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Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Write a paragraph or two (no more than 1 page) describing the misconception and then explain the correct physics by identifying the forces or physics concepts involved in simple terms, like you are talking to your 3rd grade sibling. Start with a question (will water fall out of the bucket when I stop rotating it?)Give the common answer that is incorrect (what does your intuition tell you should happen?).Explain what happens and why in simple terms (try practicing on an elementary student to make sure they understand).Use diagrams and sketches in your explanation and avoid using physics terms like rotational, centripetal, normal forces, tension, friction, acceleration, velocity, vectors, buoyancy, energy, kinetics, potential, heat capacity, etc.