Gravitational forces are stronger over shorter distances, and
weaker over longer distances. That's a big part of the reason
why our bodies are attracted to the Earth with more force than
we're attracted to Jupiter, for example.
The force doesn't just get weaker in proportion to the distance.
It gets weaker in proportion to the SQUARE of the distance.
Answer:
Angular momentum of the system is 16221465.4617 kgm²/s
Explanation:
Given that;
length of the side of the triangle L = 82 M
m = 75.0 kg × 100 = 7500 kg
distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m
effective acceleration a = 9.8 / 2 = 4.9 m/s²
we know that; effective acceleration is being provided by centripetal acceleration.
so
a = R × w²
rate of rotation w = √( a / R) = √( 4.9 / 47.34) = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
we substitute
I = 3 × 7500 × (47.34)²
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
L = 16221465.4617 kgm²/s
Therefore, Angular momentum of the system is 16221465.4617 kgm²/s
According to my research, Friction is bad when you get blisters that's when it can be harmful.
Answer:
A.) ∠CAB ≅ ∠FDE
Explanation:
The triangle SAS postulate state that two triangles that have congruent side-angle-side are congruent triangles. Take notes that the angle must be in between the two congruent sides. The congruent sides of the first triangle are CA = FD and AB=DE. The angle in between CA and AB is ∠A, while the angle between FD and DE is ∠D.
The angle needed will be:
∠A ≅ ∠D
∠CAB ≅ ∠FDE
Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ =
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² = x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m