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MArishka [77]
3 years ago
10

A python can detect thermal radiation from objects that differ in temperature from their environment as long as the received int

ensity of thermal radiation is greater than 0.60 W/m2 . Your body emits a good deal of thermal radiation. Assume that your body has a surface area of 1.6 m2 , a surface temperature of 30°C, and an emissivity e = 0.97 at infrared wavelengths. As we’ve seen, the intensity of a source of radiation decreases with the distance from the source. If you are outside on a cool, dark night, what is the maximum distance from which a python could detect your presence?
Physics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

 R = 9.92 m

Explanation:

For this exercise let's use the Stefa radiation equation

      P = σ A e T⁴

Where the Stefan-Boltzmann constant is worth 5,670 10⁻⁸ W/m²K⁴, A is the area of ​​the body and the emissivity and T is the absolute temperature

Let's calculate irradiated

     P = 5,670 10⁻⁸ 1.6 0.97 (273 +30) 4

     P = 7.417 10² W

The emitted power is distributed on a spherical surface as it progresses, whereby the intensity detected by the python I = 0.6 W / m2

        P = I A

The area of ​​the sphere is

        A = 4π R²

       

        P = I 4π R²

        R² = P / 4π I

        R = √ [7.417 102 / (4π 0.6)]

        R = √ 98.37

        R = 9.92 m

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anastassius [24]

Answer:

Option (c)

Explanation:

Both the bullets have same acceleration because they both falls under the influence of acceleration due to gravity.

The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.

the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.

8 0
3 years ago
A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi
Marianna [84]

Answer:

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0
3 years ago
What were the two different alleles for height in the pea plants that Mendel studied?
natta225 [31]
You might want to go to this website,       http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P
8 0
3 years ago
During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h
QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0
3 years ago
All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
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