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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A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s  is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

  • P = Fv................ Equation 1

Where:

  • P = work done on the horse in one second
  • F = Force of the horse
  • v = Velocity of the wagon

From the question,

Given:

  • F = 1800 N
  • v = 0.4 m/s

Substitute these values into equation 1

  • P = 1800×0.4
  • P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

Learn more about power here: brainly.com/question/25864308

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

Tony brought 9 2/3pitchers of juice to a volleyball game, and the players drank3 7/8pitchers of it. How much juice is left?

Answers

Rewrite the amounts as improper fractions:

9 2/3 = 29/3

3 7/8 = 31/8

Rewrite both fractions with a common denominator

29/3 = 232/24

31/8 = 93/24

Now subtract: 232/24 - 93/24 = 139/24

Rewrite as a proper fraction: 5 19/24

Answer 5 19/24

7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. The wire must be exactly 2.5 cm long. What two things could he do to reduce the wire's resistance

Answers

Answer and Explanation:

We know that resistance R=(\rho l)/(A)  from the given equation of resistance it is clear that resistance depends on resistivity length  and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.

So we can do two two things to reduce the resistance

  1. increase the cross sectional area
  2. decrease the resistivity of the material

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

         v² = u² + 2as

         v² = 44.50² + 2 x (-9.81) x 71.32

         v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

        v = u + at

         24.10 = 44.50 - 9.81 t , t = 2.07 s

         -24.10 = 44.50 - 9.81 t , t = 6.99 s      

Since the second ball throws after 2.3 seconds  we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

           s = ut + 0.5 at²

           71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

           u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.  

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.  

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

(a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ? 106 V/m). The plates are separated by2.98 mm and a potential difference of 5575 V is applied. (b) How close together can the plates be with this applied voltage without exceeding the breakdown strength?

Answers

Answer:

(a) 1.87×10⁶ V/m

(b) 1.12 mm closer

Explanation:

(a)

Electric Field = Electric potential/distance.

E = V/d ................... Equation 1

Where E = Electric Field, V = Electric potential, d = distance.

Given: V = 5575 V, d = 2.98 mm = 0.00298 m.

Substitute into equation 1

E = 5575/0.00298

E = 1.87×10⁶ V/m

(b)

without exceeding the breakdown strength,

make d the subject of equation 1

d = V/E.............. Equation 2

Given: E = 3×10⁶ V/m, V = 5575 V

Substitute into equation 2

d = 5575/3000000

d = 1.86 mm.

the plate will be = 2.98-1.86 = 1.12 mm closer

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up with your arms flat against your sides, and when you are standing straight up holding your arms straight out to your sides. Estimate the ratio of the moment of inertia with your arms straight out to the moment of inertia with your arms flat against your sides. (Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men's clothing stores, a man's average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches, from which an average circumference can be calculated. We'll also assume that about 20% of your body's mass is in your two arms, and that each has a length L = 1 m, so that each arm has a mass of about m = 8 kg.)

Answers

Answer:

     I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

       d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

       d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

        I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

         d = (d₁ + d₂) / 2

         d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

         r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

        M = M_man - 0.2 M_man = 80 (1-0.2)

        M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis:         Ic = ½ M r²

A rod that rotates at the end:                            I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

           I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

          D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

          I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

         I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

         I₁ = 8.258 + 5.33 + 4.129

         I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

          I₂ = ½ M r² + 2 m D²

          I₂ = ½ 64 0.508² + 2 8 0.508²

          I₂ = 8,258 + 4,129

          I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

          I₁ / I₂ = 17,717 /12,387

          I₁ / I₂ = 1.43