Answer:
ΔL = MmRgt / (2m + M)
Explanation:
The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.
ΔL = L − L₀
ΔL = Iω − 0
ΔL = ½ MR²ω
To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.
For the block, two forces act: the weight force mg downward and tension force T upward.
For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.
For the sum of forces in the -y direction on the block:
∑F = ma
mg − T = ma
T = mg − ma
For the sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Substituting gives:
mg − ma = ½ Ma
2mg − 2ma = Ma
2mg = (2m + M) a
a = 2mg / (2m + M)
The angular acceleration of the pulley is:
αR = 2mg / (2m + M)
α = 2mg / (R (2m + M))
Finally, the angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting into the previous equations gives:
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = MmRgt / (2m + M)
Answer:
Explanation:
The data indicates that point A is located midway between two charges.
To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
This field points towards Q⁻.
A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.
To find the resultant field, we add these contributions:
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
For the force acting on an electron placed at A:
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
Respuesta:
Opción e
Explicación:
La Ley de Gravitación Universal indica que toda masa puntual atrae a otra masa puntual en el universo con una fuerza que se dirige en línea recta entre los centros de masa de ambos, siendo esta fuerza proporcional a las masas de los objetos y inversamente proporcional a su separación. Esta fuerza atractiva siempre es dirigida del uno hacia el otro. La ley es aplicable a objetos de cualquier masa, sin importar su tamaño. Dos objetos grandes pueden ser considerados masas puntuales si la distancia entre ellos es considerablemente mayor que sus dimensiones o si presentan simetría esférica. En tales casos, la masa de cada objeto puede ser modelada como una masa puntual en su centro de masa.
La misma fuerza actúa sobre ambas bolas.
Response:
Clarification:
Provided
weight of disk 
diameter of disc 
weight of ring 
Force 
Overall moment of inertia
=Disc's moment of inertia +Ring's Moment of Inertia
At this point, Torque is 
Utilizing 
in this scenario
Answer:
0.018 J
Explanation:
The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as
where
represents the charge's magnitude
and 
signifies the potential difference between point P and infinity.
After substituting into the formula, we arrive at
