I believe the answer is valcanos because they release sulfur in to the air and other chemicals so it willl affect the anispher
Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,

<u>F = 51.3°</u>
<em>Convert 1nanosecond in to its SI init</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>
Explanation:
F = 20N m= m1 a=10m/s²
m=m2 a=5m/s²
F = ma
<u>for the first one</u><u>:</u><u> </u>
f=m1 × a
20 = m1 ×10
20=10m1
m1=20/10
m1=2
<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>
f=m2×a
20=m2×5
m2= 20/5
m2= 4
since F=ma
F=(m1+m2) ×a
F =(4+2)×a
F =6×a
F=20(from the question above )
20=6×a
a=20/6
a=3.33
Answer:
The new Coulomb force is q₁q₂/9πε₀r²
Explanation
The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².
Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.
Since we have air between q₂ and q₃, the coulomb force between them is
F' = q₂q₃/4πε₀d²
= q₂(q₁/κ)/4πε₀(r/2)²
= 4q₂q₁/κ4πε₀r²
= 4/κ(q₂q₁/4πε₀r²)
= 4/9 × (q₂q₁/4πε₀r²)
= q₁q₂/9πε₀r²
So, the new Coulomb force is q₁q₂/9πε₀r²