Answer:
Im actually depressed
Explanation:
jk thx for the points thoo
Answer:
A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C)
= 2.3 10³⁹
Explanation:
A) It is asked to find the force of attraction due to the masses of the particles
Let's use the law of universal attraction
F = 
let's calculate
F = 
F_g = 4.05 10⁻⁴⁷ N
B) in this part it is asked to calculate the electric force
Let's use Coulomb's law
F = 
let's calculate
F = 
F_e = 9.2 10⁻⁸N
C) It is asked to find the relationship between these forces

= 2.3 10³⁹
therefore the electric force is much greater than the gravitational force
Answer and Explanation:
Comparison between the Titan's atmosphere and earth atmosphere
- Titan atmosphere is more denser than the earth atmosphere
- The quantity of nitrogen is more in titan atmosphere than earth atmosphere titan atmosphere have about 98 % of nitrogen in the other hand earth atmosphere has only 78 %
- There are no oxygen present in titan atmosphere while in earth atmosphere it is present
- Its air is not suitable for breath but in earth atmosphere we can breath
Answer:
μ = 0.6
Explanation:
given,
speed of car = 29.7 m/s
Radius of curve = 50 m
θ = 30.0°
minimum static friction = ?
now,
writing all the forces acting along y-direction
N cos θ - f sinθ = mg
N cos θ -μN sinθ = mg

now, writing the forces acting along x- direction
N sin θ + f cos θ = F_{net}
N cos θ + μN sinθ = F_{net}

taking cos θ from nominator and denominator




now, inserting all the given values

μ = 0.6