Answer:
both experience forces or at least a force
Explanation:
it would go in the direction the other object
(second object, the one that crashed) was going
si if going right then right if left then left
plus or minus
Answer:
appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity
Explanation:
I think the correct answer would be the last option. On a cloudless day, most of the visible light that is headed towards the Earth would reach the surface wherein some is being reflected and some is being absorbed. During this time, there are no hindrances for the light so it directly reaches the surface of the Earth and is absorbed or reflected by the objects it hits.
Answer:
a = dv/ dt = -9.20 × 10⁷t + 2.55 ×10⁵
The solutio to this problem uses the concept of calculus and motion
The acceleration of the bullet is simple the result of differenyiating the velocity function with respect to time.
Explanation:
Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
![mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D_%7By%7D%5C%5Cv_%7By%7D%20%3D%20%5Csqrt%7B2gh%7D%5C%5C%3D%20%5Csqrt%7B2%20%5Ctimes%209.8%20m%2Fs%5E%7B2%7D%20%5Ctimes%201.2%7D%5C%5C%3D%204.85%20m%2Fs)
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
![v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv%5E%7B2%7D_%7Bx%7D%20%2B%20v%5E%7B2%7D_%7By%7D%7D%5C%5C%3D%20%5Csqrt%7B%286.8%29%5E%7B2%7D%20%2B%20%284.85%20m%2Fs%29%5E%7B2%7D%7D%5C%5C%3D%2011.65%20m%2Fs)
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.