Answer:
The potential energy tied to the charge diminishes.
The electric field performs negative work on the charge.
Explanation:
We will use the equations of rotational kinematics,
(A)
(B)
Here, 
and 
denote the final and initial angular displacements, respectively, whereas 
and 
represent final and initial angular velocities, and 
is the angular acceleration.
We are provided with 
and 
.
By substituting these values into equation (A), we have
Now, using equation (B),
This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.
The rod measures 450mm in length, while the disk has a radius of 75mm. An upward-supporting pin holds the assembly in place when Θ=0, and there exists a torsional spring with a constant of k=20N m/rad at the pin. One end of the rod connects to the pin, while the other connects to the disk.
The terminal velocity of the object is 12.58 m/s.
Explanation: Terminal velocity occurs when the drag force equals the weight of the object. The gravitational force is calculated as mg = 80 * 9.8 = 784 N. The drag force can then be equated to solve for the terminal velocity, resulting in v = 12.58 m/s or v = -15.58 m/s (which is unfeasible). Thus, the terminal velocity of the falling object is 12.58 m/s.
Esta pregunta está incompleta, la pregunta completa es;
La figura muestra un recipiente sellado en la parte superior por un pistón móvil. Dentro del recipiente hay un gas ideal a 1.00 atm. 20.0°C y 1.00 L.
"¿Cuál será la presión dentro del recipiente si el pistón se mueve hasta la marca de 1.60 L mientras se mantiene constante la temperatura del gas?"
Answer:
la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L manteniendo constante la temperatura del gas
Explicación:
Dado que;
P₁ = 1.00 atm
P₂ =?
V₁ = 1 L
V₂ = 1.60 L
la temperatura del gas se mantiene constante
sabemos que;
P₁V₁ = P₂V₂
por lo que hacemos la sustitución
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Por lo tanto, la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L mientras la temperatura del gas se mantiene constante
