The tension has to hold the part of the weight in the direction of the string:
T = mg*cos(theta)
Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah
complete question:
An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot
Answer:
a ≈ 5281 ft
Explanation:
The observer at the top of a 462 ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°.
The angle of depression form the top of the cliff = 5°
The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°. 90° - 5° = 85° Note sum of an angle on a right angle is 90°.
using SOHCAHTOA principle we can solve for the distance from the base of the cliff to the point on the ground(a)
tan 85° = opposite / adjacent
tan 85° = a / 462
cross multiply
462 × tan 85° = a
a = 11.4300523 × 462
a = 5280.66 ft
a ≈ 5281 ft
Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
Answer:
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Explanation:
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Answer:
695800 N/m^2 or Pa
Explanation:
Height of the water from the ground H = 71 m
Acceleration due to gravity g =9.8 m/s^2
density of water ρ= 1000 kg/m^3
The minimum output gauge pressure to make water reach height H
P= ρgH
= 1000×9.8×71= 695800 N/m^2 or Pa
