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
PV=nRT
(P)(.010)=(n)(.08201)(0)
(v1/t1)=(v2/t2)
(.010/t1)=(v2/0)
The volume would be zero
Answer:
Explanation:
r = Radius of disk = 7.9 cm
N = Number of revolution per minute = 1190 rev/minute
Angular speed is given by
The angular speed is
r = 2.98 cm
Tangential speed is given by
Tangential speed at the required point is
Radial acceleration is given by
The radial acceleration is .
t = Time = 2.06 s
Distance traveled is given by
The total distance a point on the rim moves in the required time is .
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
