Answer:
1.63 N
Explanation:
F = GMm/r^2
= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2
= 1.63 N ( 3 sig. fig.)
Newton’s first law is motion. For example, an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles
Explanation:
Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg