1 Weather-Weather is the state of the atmosphere, describing for example the degree to which it is hot or cold, wet or dry, calm or stormy, clear or cloudy
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How much money do you make?
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a Without the atmosphere, average temperature on Earth would be freezing! Below zero to be exact!
b The atmosphere protects us and animals from dangerous radiation from the Sun. Ultraviolet radiation is basically invisible rays of energy created by the sun.
cAlso, the atmosphere provides oxygen for humans and animals to breathe and C02 (Carbon dioxide) for plants.
4 warmth
5 yes it is The water on our Earth today is the same water that’s been here for nearly 5 billion years. Only a tiny bit of it has escaped out into space. As far as we know, new water hasn’t formed either.
That means there’s a very high chance the water in your glass is what thirsty dinosaurs were gulping about 65 million years ago.
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 
Answer:
Decreases to half.
Explanation:
From the question given above, the following data were obtained:
Initial mass (m₁) = m
Initial force (F₁) = F
Initial acceleration (a₁) =?
Final mass (m₂) = ½m
Final force (F₂) = ¼F
Final acceleration (a₂) =?
Next, we shall determine a₁. This can be obtained as follow:
F₁ = m₁a₁
F = ma₁
Divide both side by m
a₁ = F / m
Next, we shall determine a₂.
F₂ = m₂a₂
¼F = ½ma₂
2F = 4ma₂
Divide both side by 4m
a₂ = 2F / 4m
a₂ = F / 2m
Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:
a₁ = F / m
a₂ = F / 2m
a₂ : a₁ = a₂ / a₁
a₂ / a₁ = F/2m ÷ F/m
a₂ / a₁ = F/2m × m/F
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
From the illustrations made above, the acceleration of the car will decrease to half the original acceleration
Answer:
The force acting on a body is always equal to the product of the mass of the body and its acceleration.
Explanation:
The force of a body is defined as the product of mass and acceleration of the body.
According to Newton's second law, wherever there is a change in momentum of the body for an interval of time, there is a force acting on it.
F = (mv - mu) / t
= m (v -u) /t
= m a
Where,
(v - u)/t - is the change in velocity of the body in the interval of time. It is equal to the acceleration of the body.
Hence, the equation for the force for any body becomes, F = m x a