When he lands his horizontal velocity is
<span>28 cos40 = 21.45 m/s </span>
<span>the time in flight comes from </span>
<span>x = Vht </span>
<span>58.8 = 21.45t </span>
<span>t = 2.74 seconds </span>
<span>his vertical velocity at landing is </span>
<span>28 sin40 = -18 m/s </span>
<span>his vertical velocity equation is </span>
<span>v = V0 - gt </span>
<span>-18 = V0 - 9.81(2.74) </span>
<span>V0 = -18 + 9.81(2.74) </span>
<span>V0 = 8.88 </span>
<span>his velocity magnitude was </span>
<span>v = (8.88^2 + 21.45^2)^½ </span>
<span>v = 23.2 m/s ANSWER </span>
<span>his initial direction was </span>
<span>tanθ = 8.88/21.45 </span>
<span>θ = 22.5 degrees above the horizontal ANSWER </span>
<span>to find the time to the flight apex from launch </span>
<span>v = gt </span>
<span>8.88 = 9.81t </span>
<span>t = 0.905 s </span>
<span>in 0.905 s Eddie has risen how far above the edge </span>
<span>y = ½(9.81)(0.905^2) </span>
<span>y = 4 m </span>
<span>the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds </span>
<span>in 1.85 seconds he drops </span>
<span>y = ½(9.81)(1.85^2) </span>
<span>y = 16.7 m </span>
<span>so the height from the edge to the landing point is </span>
<span>16.7 - 4 = 12.7 m ANSWER</span>
One example of current electricity are transmission lines. These bring electricity from power stations to individual houses.
Atmosphere, geosphere, cryosphere would be the correct answer I think
Answer:
0.625 A
Explanation:
Vs = 7500 V, Is = 0.01 A
Vp = 120 V
Let the primary current be Ip.
As the transformer is ideal, so input power is equal to the output power
Vp x Ip = Vs x Is
120 x Ip = 7500 x 0.01
Ip = 0.625 A
Answer: τ = 0
Explanation:
At constant angular velocity there is no angular acceleration therefore no torque.
τ = Iα
