Respuesta :
Answer:
Approximately [tex]0.833\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
The momentum [tex]p[/tex] of an object of mass [tex]m[/tex] moving at a velocity of [tex]v[/tex] is:
[tex]p = m\, v[/tex].
Under the assumptions, momentum in the horizontal direction should be conserved. In other words, the following should be equal:
- Momentum of the boat before the rain started;
- Momentum of the boat and water accumulated in the boat, combined, after the rain.
The velocity of the boat after the rain can be found by dividing the total momentum by the combined mass. The initial momentum of the boat before the rain can be found by taking the product between the mass of the boat and the initial velocity of the boat.
At [tex]1.00\; {\rm m\cdot s^{-1}}[/tex], the momentum of the [tex]250\; {\rm kg}[/tex] boat before the rain is:
[tex](250\; {\rm kg})\, (1.00\; {\rm m\cdot s^{-1}}) = 250\; {\rm kg\cdot m\cdot s^{-1}}[/tex].
After the rain, it is given that the amount of water collected in the boat is [tex](100)\, (0.5)\; {\rm kg} = 50\; {\rm kg}[/tex]. The combined mass of the boat and the water collected in it would be [tex](250\; {\rm kg}) + (50\; {\rm kg}) = 300\; {\rm kg}[/tex].
The value of momentum stays the same. Divide momentum by the new mass of the boat to find the new velocity:
[tex]\displaystyle \frac{250\; {\rm kg\cdot m\cdot s^{-1}}}{300\; {\rm kg}} \approx 0.833\; {\rm m\cdot s^{-1}}[/tex].
In other words, under the assumptions, the new velocity of the boat would be [tex]0.833\; {\rm m\cdot s^{-1}}[/tex].
Final answer:
The question involves calculating the time for a boat to travel downstream and upstream, determining the heading angle to cross straight over a river, and finding the resultant velocity of a boat in a crosscurrent. Physics principles such as relative velocity, vector addition, and dynamics are applied.
Explanation:
The question deals with the effects of external forces on a boat's motion and uses concepts of relative velocity, vector addition, and dynamics within the study of physics. Here are some solutions to the situations described:
- Rowing 1.5 km downstream with an 8.0 km/h boat speed in still water and a 3.0 km/h river flow would result in a resultant velocity of 11.0 km/h (since the river's current aids in the boat's movement). To find the time required, divide the distance (1.5 km) by the resultant velocity (11.0 km/h), resulting in approximately 0.136 hours or about 8.18 minutes.
- For the return trip (rowing upstream), the river’s current works against the boat, so subtract the river’s speed from the boat's speed in still water to get 5.0 km/h. Thus, the time taken to travel 1.5 km upstream would be the distance divided by this slower speed, which would take about 0.3 hours or 18 minutes.
- The direction to aim the boat straight across the river would involve a heading angle that compensates for the river's current. This will depend on the specific speeds of the river and the boat's ability to move in still water.
- If the river is 0.8 km wide and the boat attempts to move straight across with a speed of 3.8 m/s while the river’s current is 6.1 m/s, the velocity of the boat with respect to Earth can be found by constructing and solving a vector triangle or using Pythagorean theorem and trigonometry to determine the resultant velocity and the angle made with the shore.
The subject matter requires an application of the principles of physics to solve practical problems involving motion, forces, and vectors. This is typically covered in high school physics courses.
