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dimulka [17.4K]
3 years ago
13

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ce it travels each succeeding second would be:_________
a) constant.
b) less and less each second.
c) greater than the second before
Physics
1 answer:
ioda3 years ago
5 0

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

h=ut+\frac{1}{2}gt^2

here u=0 and t=time taken

h=\frac{1}{2}gt^2

for t=1 s

h_1=\frac{1}{2}g

for t=2 s

h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g

distance traveled in 2 nd sec=2g-\frac{1}{2}g=\frac{3}{2}g

for t=3 s

h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g

distance traveled in 3 rd sec=\frac{9}{2}g-2g=\frac{5}{2}g

so we can see that distance traveled in each successive second is increasing

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Example of what happens to particles when their kinetic energy decreases?
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Explanation:

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3 years ago
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An immense body of air characterized by similar properties at any given altitude is known as _____.
sertanlavr [38]

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Explanation:

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3 years ago
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a. What is the peak wavelength for AM0? What temperature T corresponds to this peak wavelength for a blackbody source? Assuming
atroni [7]

Answer:

A) T = 5510 K , B)   I = 5,226 10⁷ W / m² , C)   I₂ = 1128 W / m²

Explanation:

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      Am = 1 / cos θ

The AM0 value corresponds to solar radiation in the outer part of the Earth's atmosphere.

The peak of this emission is the peak that emitted from the sun

       λ = 526 nm

To find the temperature that corresponds to this emission we use the Wien displacement law

       λ T = 2,898 10⁻³

      T = 2,898 10⁻³ / 526 10⁻⁹

      T = 5510 K

i) The radiance on the surface of the sun is

           I = P / A

We can calculate the potency by Stefan's law, for a black body

         P = σ A e T⁴

         P / A = σ e T⁴

The σ constant is value 5,670 10⁻⁸ W / m²K⁴, we will assume that the Sun emits as a black body, so e = 1

            I = sig T⁴

            I = 5,670 10⁻⁸  5510⁴

            I = 5,226 10⁷ W / m²

ii) the irradiation at a distance of 1 ua (1,496 1011 m)

Let's use the relationship

           P = I A

           I₁ A₁ = I₂ A₂

           I₂ = I₁ A₁ / A₂

 

The area of ​​a sphere is

          A = 4π R²

Let's replace

         I₂ = I₁ (r₁ / r₂)²

  Index 1 corresponds to the sun and the index to Earth that is an astronomical unit

        r₁ = 6.96 10⁸m        (Sun radius)

        r₂ = 1,498 1011 m    (Earth-Sun distance)

Calculous

       I₂ = 5,226 10⁷ (6.96 10⁸ / 1,498 10¹¹)²

       I₂ = 1.1281 10³ W / m²

       I₂ = 1128 W / m²

8 0
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Answer:

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In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

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