In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Explanation:
The given data is as follows.
height (h) = 98.0 m, speed (v) = 73.0 m/s,
Formula of height in vertical direction is as follows.
h =
,
or, t =
Now, formula for the required distance (d) is as follows.
d = vt
=
=
= 326.5 m
Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.
Answer:
HCl, Option A.
Explanation:
Hcl is a strong acid, NaOH is a strong base, the saliva has a relatively neutral pH and blood is slightly acidic.
Answer:beats
Explanation:l got it right gang