<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
Answer:
the inertia provides the tendency to maintain speed and keep moving
Answer:
5,970 N
Explanation:
m = 597 kg
a = 10 m/s^2
Plug those values into the following equation:
F = ma
F = (597 kg)(10 m/s^2)
F = 5,970 N
Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>
<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>
