A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
1 answer:
The work done on the ship is
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and of the displacement
In this problem, we have:
(force acting on the ship)
d = 3.00 km = 3000 m (displacement of the ship)
(because the force is horizontal, and the displacement is horizontal as well)
Therefore, the work done on the ship is
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If this case could ever happen, the speed would follow from this formula:
with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."
The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)
Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
Then, the time of the second phase will be:
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
So, the total distance covered by this object in meters will be the sum of all the distances we found: