Answer:
v = 10 m/s
Explanation:
Given that,
Distance covered by a sprinter, d = 100 m
Time taken by him to reach the finish line, t = 10 s
We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,
v = d/t
Hence, his average velocity is 10 m/s.
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.
![E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DF%2Ad%5C%5C400%20%3D%208000%2Ad%5C%5Cd%20%3D%200.05%20%5Bm%5D%20%3D%2050%20%5Bmm%5D)
Both are constants used in the definition of Forces (gravitational and electric,respectively)
Since those constants are proportional to the magnitude of the forces:
Having a small gravitational constant explains why there is no apparent force of attraction with objects of considerable low mass (they would need to have great value of mass for the equation to give an apreciable force)
Electrical interactions are usually strong, and thus require an appropiate constant to depict the phenomenon. We deal in this case with charges really small, but the forces are in different order of magnitude.
Answer:
The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.
Explanation:
Answer:
Explanation:
velocity of first projectile after 3 s
v = u - gt
v = 49.4 - 9.8 x 3
= 20 m /s
Velocity of second projectile after 3 s after being dropped from rest
v = u + gt
= 0 + 9.8 x 3
= 29.4 m /s
They will be moving in opposite direction at the time of meeting , so their relative velocity
= 20 + 29.4 = 49.4 m /s
From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .
