Answer:
A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?
According to Dalton's law of partial pressure, the total pressure exerted is simply equal to the sum of the partial pressures of the individual gases. Given that all three samples of gas each exert 740 mmHg, when they are placed in a single 2 L container, they exert a pressure of 2220 mmHg on the container which is the sum of their individual pressures.
Answer:
y = 60 mph
Explanation:
given,
average speed of Sarah driving from Orlando to Tampa = 30 mph
average speed of return = ?
whole trip average speed = 40 mph
now,
![avg\ speed = \dfrac{total\ distance}{total\ time}](https://tex.z-dn.net/?f=avg%5C%20speed%20%3D%20%5Cdfrac%7Btotal%5C%20distance%7D%7Btotal%5C%20time%7D)
Let x be the distance from Orlando to Tampa
y be the speed of return
![avg\ speed = \dfrac{x + x}{\dfrac{x}{30}+\dfrac{x}{y}}](https://tex.z-dn.net/?f=avg%5C%20speed%20%3D%20%5Cdfrac%7Bx%20%2B%20x%7D%7B%5Cdfrac%7Bx%7D%7B30%7D%2B%5Cdfrac%7Bx%7D%7By%7D%7D)
![avg\ speed = \dfrac{2}{\dfrac{1}{30}+\dfrac{1}{y}}](https://tex.z-dn.net/?f=avg%5C%20speed%20%3D%20%5Cdfrac%7B2%7D%7B%5Cdfrac%7B1%7D%7B30%7D%2B%5Cdfrac%7B1%7D%7By%7D%7D)
![40= \dfrac{2\times 30 y}{30+y}](https://tex.z-dn.net/?f=40%3D%20%5Cdfrac%7B2%5Ctimes%2030%20y%7D%7B30%2By%7D)
60 + 2 y = 3 y
y = 60 mph
hence, speed of the return is equal to 60 mph.
Answer:
i am not a point taker i help people when they need help
Explanation:
3
i think it is right
Answer:
11.48 degree N of W
Explanation:
We are given that
Wind velocity=
km/h
Because wind is blowing towards south
Air speed=![v_a=-212km/h](https://tex.z-dn.net/?f=v_a%3D-212km%2Fh)
Because the captain want to move with air speed in west direction.
x component of relative velocity=-212 km/h
y-Component of relative velocity=-43km/h
Direction=![\theta=tan^{-1}(\frac{y}{x})](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7By%7D%7Bx%7D%29)
N of W
Hence, the direction in which the pilot should set her course to travel due west=11.48 degree N of W