Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm
Divide 56 by 60 then x the answer by 6 then you get 5.6km which is your answer.
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