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3 years ago

What two things does the wavelength of peak of the blackbody

Physics

1 answer:

Georgia [21]3 years ago

8 0

Answer:

The correct option is;

c. Temperature and color

Explanation:

Based on Wien's Law which can be presented mathematically as follows;

$\lambda _{max} = (0.29 \ cm \ K) / T$

The wavelength of the peak of the blackbody curve tells the temperature and color.

An example is the Sun that has a temperature of 5800 K, hence the peak wavelength is given by $\lambda _{max} = (0.29 \ cm \ K) / 5800 = 0.00005 \ cm$ = 500 nano-meter from which the color can be deduced as 500 nano-meter is the wavelength of yellow light.

You might be interested in

These are equal and opposite forces that do not cause a change in position or motion. True or False

Contact [7]

Answer:

TRUE

Explanation:

Balance forces are usually defined as the two distinct force that acts on an object but in opposite directions. These two acting forces are equal in size or magnitude. When this type of force is applied on any object, it signifies that the object is stationary or it is moving at a constant speed and in the same direction.

This force is comprised of two most important properties namely the strength and direction. When any of the two forces is higher then it result in the motion of the object.

Thus, the above given statement is TRUE.

3 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to posit

anyanavicka [17]

At point E

the kinetic energy of the rollercoaster is small compared to the potential energy
the potential energy is greater than the kinetic energy
the total energy is a mixture of potential and kinetic energy

<h3>What is the energy of the roller coaster at point E?</h3>

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

the kinetic energy of the rollercoaster is small compared to the potential energy
the potential energy is greater than the kinetic energy
the total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

Learn more about potential and kinetic energy at: brainly.com/question/18963960

#SPJ1

5 0

2 years ago

brainly an andean condor with a wingspan and a mass soars along a horizontal path. model its wings as a rectangle with a width.

Anna35 [415]

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

<h3>Difference in pressure between the top and bottom of the wingspan</h3>

The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

<h3>Area of the wingspan</h3>

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

<h3>Weight of the Andean condor</h3>

W = mg

W = 9 x 9.8

W = 88.2 N

<h3>Pressure at the top surface of condor's wings</h3>

The pressure at the top surface of condor's wings is due to atmospheric pressure

P(top) = 14.7 Psi = 101,325 Pa

<h3>Pressure at the bottom surface of condor's wings</h3>

The pressure at the bottom surface is due to weight of andean condor.

P = W/A

P(bottom) = 88.2/0.729

P(bottom) = 120.99 Pa

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is calculated as;

ΔP = P(top) - P(bottom)

ΔP = 101,325 Pa - 120.99 Pa

ΔP = 101,204 Pa

The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

Learn more about pressure here: brainly.com/question/25736513

5 0

2 years ago

Scenario

Anvisha [2.4K]

Answer:

1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

0 = y₀ + 0 - ½ g t2

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 2650/9.8)

t = 23.26 s

this is the horizontal scrolling time

x = v₀ t

x = 366.67 23.26

x = 8527 m

the speed at the point of arrival is

v_y = v_{oy} - g t = 0 - gt

v_y = - 9.8 23.26

v_y = -227.95 m / s

Module and angle form

v = $\sqrt{v_x^2 + v_y^2}$

v = √(366.67² + 227.95²)

v = 431.75 m / s

θ = tan⁻¹ (v_y / vₓ)

θ = tan⁻¹ (227.95 / 366.67)

θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

cos θ = v₀ₓ / v₀

sin θ = v_{oy} / v₀

v₀ₓ = v₀ cos θ

v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + vo sin θ t - ½ g t²

0 = -50 + vo sin 50 t - ½ 9.8 t²

x = x₀ + v₀ₓ t

0 = x₀ + vo cos θ t

0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

50 = 0,766 vo t – 4,9 t²

400 = 0,643 vo t

resolved

50 = 0,766 ( $\frac{400}{0.643 \ t}$ ) t – 4,9 t²

50 = 476,52 t – 4,9 t²

t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

t = [97.25 ± $\sqrt{97.25^2 - 4 \ 10.2}$ ] / 2

t = 97.25 ±97.04] 2

t₁ = 97.145 s

t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

t = 97.145 s

let's find the initial velocity

x = x₀ + v₀ cos 50 t

0 = -400 + v₀ cos 50 97.145

v₀ = 400 / 62.44

v₀ = 6.4 m / s

5 0

3 years ago