The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
acting against the motion. Since their resultant must be zero, we have:
The frictional force is
where
is the coefficient of kinetic friction
is the weight of the block.
Substituting these values, we find the magnitude of the force F:
At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle. Then he brings the bottle down to the base of the mountain.
The pressure on the outside of the bottle is greater than it was when he put the cap on. If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.
The water is pretty much unsquishable.
Biut the air in there can be <em>COMPRESSED</em>. The air gets squished into a smaller space, and the bottle wrinkles in slightly.
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
The answer would be 5.6x10^5
Answer:
.067 so C
Explanation:
I asked my sister who is in 2nd grade and she said it was right so you are good! =). have a great day!
