F=mgsinα –F(fr)=
= mgsinα - μmgcosα
a=F/m =g(sinα – μcosα)
=9.8(sin14-0.57•cos14)= -3.04 m/s²
v²=v₀²+2ad
v=sqrt{ v₀²+2ad} =
=sqrt{18²+2• (-3.04) •25}=13.1 m/
You finna have to pay me lol what it look like though and i’ll see if it’s worth it make some
Answer:
A) 11.28 x 10^(7) A.m²
B) 2.258 x 10^(17)A
Explanation:
A) The current density is given by the formula ;
J = nqv
Where n is the density of protons in the solar wind which is 12.5 cm³ or 12.5 x 10^(-6) m³
q is the proton charge which is 1.6 x 10^(-19) C
v is velocity which is 564km or 564000m
Thus, J = 12.5 x 10^(-6) x 1.6 x 10^(-19) x 564000 = 11.28 x 10^(7) A.m²
B) the formula for the total current the earth received is given as;
I = JA
The effective area is the cross section of the earth and thus,
Area = πr² where r is the radius of the earth given as: 6.371 x 10^(6)
A = π(6.371 x 10^(6)) ²
So I = 11.28 x 10^(7) x π(6.371 x 10^(6))² = 2.258 x 10^(17)A
