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callieroberts1699
18.03.2020 •
Physics
An rv travels 45 km east and stays the night at a KOA. The next day it travels for 3 hours to the north l, traveling 110 km. What is the displacement over the two days for the RV?
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Ответ:
The displacement of the rv over the two days is 118.85 km.
The given parameters;
initial displacement of the rv = 45 km eastfinal displacement of the rv, = 110 km northThe displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;
Thus, the displacement of the rv over the two days is 118.85 km.
Learn more about displacement here: link
Ответ:
The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.
Explanation:
Given:
Distance moved in the East direction (d) = 45 km
Distance moved in the North direction (D) = 110 km
Displacement is defined as the difference of final position and initial position.
Let us draw a diagram representing the above situation.
Point A is the starting point and point C is the final position of RV.
So, the displacement of RV in two days is given as:
Displacement = Final position - Initial position = AC
Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.
Using Pythagoras theorem, we have:
Plug in the given values and solve for AC. This gives,
Now, the direction of displacement with the east direction is given as:
Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.
Ответ:
cool photos can they compete tho??
Explanation: