PHYSICS COLLEGE

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer 1

Answer:

Answer:

The number of available energy states per unit volume is $4.01*10^(48)$

Explanation:

Given that,

Average energy $E=2*10^(-4)\ eV$

Photon = $4*10^(-5)\ eV$

We need to calculate the number of available energy states per unit volume

Using formula of energy

$g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)$

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

$g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)$

$g(\epsilon)d\epsilon=4.01*10^(48)$

Hence, The number of available energy states per unit volume is $4.01*10^(48)$

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So to deal with the irrational belief in REBT, we must Group of answer choices

A. Consult with a friend and get their feeback

B. Dispute the beliefs by asking if these are true and examining the evidence

C. Seek mental health counseling

D. It is just too hard so let's just forget it.

Answers

Answer:

i believe the answer is B

Explanation:

Seeking the right answer is the best thing to do

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф= $((wi+wf)/(2) )$ t

Ф= $((0+2)/(2) )5=5$ rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

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

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

$d = \theta R$

$d = (\omega + \omega_o)(1)/(2)t R$

$d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )$

$d = 4\pi (1.75)$

$d = 7\pi \texttt{ m}$

$d \approx 22 \texttt{ m}$

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Learn more

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$

x = 0.15 meter

Answer:

Explanation:

Force constant of spring K = 21 N /m

we shall find the common velocity of putty-block system from law of conservation of momentum .

Initial momentum of putty

= 5.3 x 10⁻² x 8.97

= 47.54 x 10⁻² kg m/s

If common velocity after collision be V

47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V

V = .937 m/s

If x be compression on hitting the putty

1/2 k x² = 1/2 m V²

21 x² = ( 5.3x 10⁻² + .454) x .937²

x² = .0212

x = .1456 m

14.56 cm

An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an instant, exceed 16 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit?A)8 sqrt 2 V

B) 16 sqrt 2 V

C) 256 V

D) 8

Answers

Answer:

A) $V_(rms)=8√(2) V$

Explanation:

Maximum voltage = $V_(max)=16 V$

Maximum voltage and rms voltage are related to each other by

$V_(max)=V_(rms) * √(2) \nV_(rms)=(V_(max))/( √(2))\nV_(rms)=(16)/(√(2)) \nV_(rms)=8√(2) V$

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

In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m