Answer:
I do I make a brinliest can you please can me
Answer:
Most of the oxygen in the atmosphere is in the form of O2, which means it is made up of molecules containing two oxygen atoms. Ozone, however is O3, which means it is made up of molecules containing three oxygen atoms. O2 is what we breath, and what plants release from photosynthesis. Ozone occurs naturally high in the stratosphere, where it absorbs ultraviolet light, protecting us here on the surface from skin cancer. Ozone can also occur closer to the surface of the earth as a pollutant. It is formed from reactions with O2 and chemicals emitted from factories and cars. It comes in the form of smog.
So in general:
Oxygen (O2): Essential to human life
Ozone (O3) in the stratosphere: essential to protecting life on earth
Ozone (O3) on the surface of the earth: toxic to human life, caused by pollution
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.