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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 15.0 m long plane is inclined at 30.0 degrees. If the coefficient of friction is 0.426, what force is required to move a 40.0
FromTheMoon [43]

consider the motion of the mass parallel to the incline

v₀ = initial velocity at the bottom of incline = 0 m/s

v = final velocity at the top of incline = 8.00 m/s

a = acceleration

d = displacement = L = length of incline = 15 m

using the equation

v² = v²₀ + 2 a d

8² = 0² + 2 a (15)

64 = 30 a

a = 64/30

a = 2.13 m/s²

F = applied force

from the force diagram, perpendicular to incline , force equation is given as

N = mg Cos30

μ = Coefficient of friction = 0.426

frictional force acting on the mass is given as

f = μ N

f = μ mg Cos30

parallel to incline , force equation is given as

F - f - mg Sin30 = ma

F - μ mg Cos30 - mg Sin30 = ma

inserting the values

F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)

F = 425.82 N

4 0
3 years ago
When light hits a RED WAGON some of the energy is absorbed and _______.
Gnoma [55]
Blue light is absorbed
7 0
3 years ago
Read 2 more answers
If a 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft
mamaluj [8]

Answer: 3217.79 hours.

Explanation:

Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).

Power = 0.4 watt

Mass of climber = 140 lb

= 140 x 0.4535 kg  [∵ 1 lb= 0.4535 kg]

⇒ Mass of climber (m) = 63.50 kg

Let h_1=29,029\ ft= 8848.04\ m\ \ \ \  [ 1 ft=0.3048\ m ] and h_2= 4,600 ft = 1402.08\ m

Now, Energy saved =mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J

\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}

Hence, she can power her 0.4 watt flashlight for 3217.79 hours.

5 0
4 years ago
charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³  N / C .

3 0
3 years ago
One hypothesis states that plate movement results from convection currents in the
brilliants [131]
I believe is A) Inner core
8 0
3 years ago
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