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Tamiku [17]
3 years ago
11

Make a list of possible consequenes of solar flares to the following members of your community

Physics
1 answer:
SashulF [63]3 years ago
4 0
The Majestic Aurora


Nature's Charming Rainbows 

Sunrise, Sunset, and Other Atmospheric Phenomena

Global Warming

Nature



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belka [17]
Both mass and weight
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3 years ago
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"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front
cluponka [151]
If i wouldve know i would tell you sorry
4 0
3 years ago
Are the two atoms shown below in the image from the same element?
kvasek [131]

The answer is yes

They are isotopes.

7 0
3 years ago
A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i
klio [65]

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer ,  so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. =  λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

5 0
3 years ago
The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
3 years ago
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