<u>Given that:</u>
Ball dropped from a bridge at the rate of 3 seconds
Determine the height of fall (S) = ?
As we know that, S = ut + 1/2 ×a.t²
u =initial velocity = 0
a= g =9.81 m/s (since free fall)
S = 0+ 1/2 × 9.81 × 3²
<em> S = 44.145 m</em>
<em>44.145 m far is the bridge from water</em>
Answer:the 90% is spent performing other functions
Explanation:
Answer:
The greater the amplitude the greater the energy.
(Think of a water wave - which carries greater energy a 1 ft wave or
a 10 ft wave)
Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]
(d) Acceleration is a vector quantity
