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g100num [7]
3 years ago
6

If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? Wh

at forms of radiation might penetrate the clouds and reach the ground?
Physics
1 answer:
Fudgin [204]3 years ago
4 0

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

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A 140 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless trac
photoshop1234 [79]

Answer:

Explanation:

mass attached m = .14 kg

force constant k = 5N / m

displacement

= amplitude of oscillation

A = .03 m

A ) period of motion = 2\pi\sqrt{\frac{m}{k} }

= 2 x 3.14 \sqrt{\frac{.14}{5} }

T = 1.05 s

B ) maximum speed of block = angular velocity x amplitude

= (2π /T)  x A

= (2 x 3.14 x .03) / 1.05

= .1794 m / s

17.94 cm /s

C )

maximum acceleration = angular velocity² x amplitude

= (2π /T)² x A

= (2π /1.05)² x .03

= 1.073 m / s²

D )

position

S = A cos ωt , ω is angular velocity

S = .03 cos(2πt /T)

= .03 cos 5.98 t

v =ω A sin(2πt /T)

= 5.98 x .03 sin5.98t

= .1794 sin5.98t

acceleration = ω²A sin5.98t

= 1.073 sin5.98t

7 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
shtirl [24]

Answer:

\lambda=1.39\times 10^{-4}\ s^{-1}

Explanation:

Given that,

The half-life of Barium-139 is 4.96\times 10^3

A sample contains 3.21\times 10^{17} nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}

Put all the values,

\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}

So, the decay constant is 1.39\times 10^{-4}\ s^{-1}.

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inessss [21]
Water vapor decreases
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olganol [36]

Answer:

true

Explanation:

5 0
3 years ago
HELP THIS IS DUE IN 2 DAYS PLS (the question in this attachment)
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