just a couple more questions
1.) The power needed to accelerate a projectile from rest to is launch speed v in a time t is 43.0 W. How much power is needed to accelerate the same projectile from rest to launch speed of 2v in a time of 1/2t?
2.)A ball is fixed to the end of a string , which is attached to the eiling at point P, which is at the top. the ball is positioned below it, what enables it to go back to point P? Hope this makes sense. Like swinging the ball basically
ive been looking for the same one
1.) To find the power needed to accelerate the projectile from rest to 2v in a time of 1/2t, we can use the concept of kinetic energy and work done.
The power is defined as the rate at which work is done or energy is transferred. In this case, the work done is equal to the change in kinetic energy of the projectile.
The initial kinetic energy of the projectile is zero since it is at rest. The final kinetic energy is given by (1/2)mv^2, where m is the mass of the projectile and v is its launch speed.
The work done on the projectile is equal to the change in kinetic energy, which is (1/2)mv^2 - 0 = (1/2)mv^2.
The time taken to accelerate the projectile to 2v is 1/2t, which means we need to find the power for this time interval.
The formula for power is given by P = W/t, where P is the power in watts, W is the work done in joules, and t is the time interval in seconds.
Substituting the values we obtained earlier, we have:
P = (1/2)mv^2 / (1/2t)
Simplifying, we get:
P = (mv^2) / t
Since the power needed to accelerate the projectile to v is given as 43.0 W, we have:
43.0 = (mv^2) / t
Now, to find the power needed to accelerate the projectile to 2v in a time of 1/2t, we substitute 2v for v and 1/2t for t:
P' = (m(2v)^2) / (1/2t)
Simplifying, we get:
P' = 4(mv^2) / (1/2t)
P' = 8(mv^2) / t
Notice that the expression 8(mv^2) / t is four times the expression (mv^2) / t. Therefore, the power needed to accelerate the projectile to 2v in a time of 1/2t is four times the power needed to accelerate the projectile to v. Thus, the power needed is:
P' = 4 * 43.0 W
P' = 172.0 W
Therefore, the power needed to accelerate the same projectile from rest to a launch speed of 2v in a time of 1/2t is 172.0 W.
2.) In order for the ball attached to the string to swing back and forth, it needs an initial force or push to move it away from the equilibrium position. Once the ball is displaced from the equilibrium, several forces act upon it.
As the ball swings, it experiences two primary forces: tension and gravity. The tension force is provided by the string, which keeps the ball in circular motion. The gravitational force acts vertically downward towards the center of the Earth.
As the ball swings towards the highest point, tension provides the centripetal force necessary to keep it moving in a circular path. At that point, when the ball reaches its highest position, the tension force is at its maximum, counteracting the gravitational force.
When the ball starts moving back towards the original position, the force of gravity becomes the predominant force, pulling the ball downwards. As it accelerates downward, the string changes from being nearly horizontal to oblique, providing a combined force along the circular motion and the downward gravitational force.
This back and forth movement continues until the ball eventually comes to a stop at its lowest point. At this point, the tension force becomes zero, and the gravitational force is the only force acting on the ball.
In summary, the initial force or push applied to the ball sets it in motion, and the tension force from the string keeps it moving in a circular path. The interplay between the tension force and gravitational force allows the ball to swing back and forth, completing its motion.