How many chicken feet hatch from 3 dozen eggs? Explain.

sineoko [7]

Answer:

a) λ = 5,084 10⁻⁷ m , b) P = 3.63 10²⁶ W , c) P = 5.8 10²⁷ W and d) λ = 2.54 10⁻⁷ m

Explanation:

a) The maximum emission of the sun can be calculated using the Win equation

λ T = 2,898 10⁻³ m.K

λ = 2,898 10⁻³ / T

λ = 2,898 10⁻³ / 5700

λ = 5,084 10⁻⁷ m

λ = 5,084 10⁻⁷ m (1 10⁹ nm / 1m) =

λ = 5,084 10² nm = 508.4 nm

photon in the visible range

b) The emission of the Sun, is described by the Stefan equation

P = σ A e T⁴

Where σ is the Stefan-Boltzmann constant that vslue is 5,670 10-8 W/m²K⁴, A area of the Sun, and e the emissivity that for a perfect black body is 1

In order to use this equation, we must calculate the area of the sun, we consider it a perfect sphere

r = 695,000 km (1000m / 1 km) = 6.95 10⁸ m

Area of a sphere

A = 4π R²

A = 4π (6.95 10⁸8)²

A = 6.07 10¹⁸ m²

P = 5,670 10⁻⁸ 6.07 10¹⁸ 1 5700⁴

P = 3.63 10²⁶ W

c) The new temperature is double the previous one

T = 2 To

Let's substitute in the formula and calculate

P = σ A e (2To)⁴

P = σ A e T⁴ 2⁴

Po = σ A e T4 = 3.63 10 26 W

P = 16 Po

P= 16 (3.63 10²⁶)

P = 5.8 10²⁷ W

d) Let's calculate the explicit value of the temperature and use the Win equation

T = 2 5700

T = 11400K

λ = 2,898 10⁻³ / 11400

λ = 2.54 10⁻⁷ m

λ = 2.54 10²nm = 254 nm

photon in the UV range

5 0

3 years ago

If a galaxy has an apparent radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy

ddd [48]

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

3 0

3 years ago

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a

s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0

3 years ago

Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.

Julli [10]

Answer:

(a) the total momentum of the system before the collision = -2m kg.m/s.

(b) the total momentum of the system after the collision = -2m kg.m/s.

(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)

Explanation:

Given;

mass of Puck 1 , = m

mass of Puck 2, = m (since they have the same mass m)

initial velocity of Puck 1, u₁ = 10 m/s to the left

initial velocity of Puck 2, u₂ = 8 m/s to the right

Let the rightward direction be positive direction

Let the leftward direction be negative direction

(a) the total momentum of the system before the collision;

P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)

P₁ = (-mu₁) + mu₂

P₁ = mu₂ - mu₁

P₁ = m(u₂ - u₁)

P₁ = m(8 - 10)

P₁ = -2m kg.m/s

(b) the total momentum of the system after the collision;

Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.

Thus, the total momentum of the system after the collision is -2m kg.m/s.

(c) puck 1's velocity after the collision in component form

$v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s$

8 0

3 years ago

A ball was thrown from a projectile building of 30m which moves at a constant velocity of 20m/s and has an angle of 30degrees to

qwelly [4]

Answer:

Time of flight = 4.08seconds

Horizontal component of initial velocity is 17.32m/s

Explanation: complete question( and the horizontal component of the initial velocity.)

The equation for time of flight of a projectile is given as T= 2u/g

T=( 2×20)/9.8

T= 40/9.8= 4.08seconds

Horizontal component of initial velocity Vix= Vi Costheta

Vix= 20× cos 30°

Vix= 20×0.8660

Vix= 17.32m/s

6 0

3 years ago

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