Si tenemos dos placas paralelas y un campo electrico de 8kN/C entre ellas y además conocemos la separación de estas placas que es igual a 0.03m. Entonces, cuando una carga se mueve de un punto a a un punto b el trabajo que es realizado sobre la carga por el campo eléctrico viene dado por la siguiente ecuación:
Dado que la carga se mueve desde la placa negativa hasta la placa positiva, entonces:
Finalmente, sustituyendo:
Actually Welcome to the Concept of the Force.
Force = Mass * Acceleration.
here, Mass = 45 kg and Force = 65 N
hence, Acceleration = 65/45
===> Acc. = 1.44 m/s^2
hence the acceleration is 1.44 m/s^2
Given data:
* The mass of the book is 2 kg.
* The initial height of the book is 0.78 m.
* The final height of the book is 2.1 m.
Solution:
(A). The work done by the gravity on the book is,
where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,
The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.
Substituting the known values,
Thus, the work done by the gravitational force is -25.9 J.
(B). The work done by the hand on the moving the book is,
Thus, the work done by the hand on the book is 25.9 J.
Answer:
a. b. D = v₀²/2g
Explanation:
Here is the complete question
A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance
Find the time at which the two balls collide.
Express your answer in terms of the variables D ,v₀ , and appropriate constants..
t = ?!
Part B
Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.
Express your answer in terms of the variables v₀ and g .
D =?!
Solution
The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.
The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.
The two balls collide when their vertical distances are equal. That is H₁ = H₂
So, D - v₀t + gt²/2 = v₀t - gt²/2
Collecting like terms
D + gt²/2 + gt²/2 = v₀t + v₀t
D +gt² = 2v₀t
gt² - 2v₀t + D = 0.
Using the quadratic formula,
B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.
0 = v₀² - 2gs
s = v₀²/2g.
Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,
0 = v₀ - gt₀
t₀ = v₀/g
Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.
So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g