Answer:
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Explanation:
First, we must calculate the resultant force (
), in newtons, by vectorial sum:
![\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20%5B%28-200%5C%2CN%29%5Ccdot%20%5Ccos%2060%5E%7B%5Ccirc%7D%2B%28400%5C%2CN%29%5Ccdot%20%5Ccos%2045%5E%7B%5Ccirc%7D%2B300%5C%2CN%5D%5C%2C%5Chat%7Bi%7D%20%2B%20%5B%28200%5C%2CN%29%5Ccdot%20%5Csin%2060%5E%7B%5Ccirc%7D%20%2B%20%28400%5C%2CN%29%5Ccdot%20%5Csin%2045%5E%7B%5Ccirc%7D-100%5C%2CN%5D%5C%2C%5Chat%7Bj%7D)
(1)
Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:
Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:
Where 
is the direction of the resultant force, in sexagesimal degrees.
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Answer:
<em>20 m/s in the same direction of the bus.</em>
Explanation:
<u>Relative Motion
</u>
Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.
If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.
Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3
Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all
Answer: waves transport energy, not water. As a wave crest passes, the water particles move in circular paths. The movement of the floating inner tube is simulacra to the movement of the water particles. Water particles rise as a wave crest approaches.
Explanation:
