Answer:
Explanation:
Let the vertical height by which it descends be h . Let it acquire velocity of v .
1/2 mv² = mgh
v² = 2gh
As it leaves the surface of sphere , reaction force of surface R = 0 , so
centripetal force = mg cosθ where θ is the angular displacement from the vertex .
mv² / r = mg cosθ
(m/r )x 2gh = mg cosθ
2h / r = cosθ
cosθ = (r-h) / r
2h / r = r-h / r
2h = r-h
3h = r
h = r / 3
Washington DC and new Mexico
Answer:
Velocidad = 2 m/s
Explanation:
Dados los siguientes datos;
Distancia = 80 m
Tiempo = 40 s
Para encontrar la velocidad del automóvil;
La velocidad se puede definir como la tasa de cambio en el desplazamiento (distancia) con el tiempo.
La velocidad es una cantidad vectorial y, como tal, tiene magnitud y dirección.
Matemáticamente, la velocidad viene dada por la ecuación;
Sustituyendo en la fórmula, tenemos;
Velocidad = 80/40
Velocidad = 2 m/s
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.