Math
looks good
Physics
horizontal velocity (Vh) is the velocity (V) times the cosine of 20.0º vertical velocity (Vv) is V times the sine of 20.0º or __ Vv = Vh * tan(20.0º) = (104/4.60) * tan(20.0º) = 8.23 when the stone hits the ground, H = 0 0 = (-.5 * g * 4.60^2) + (8.23 * 4.6...
Physics
a) vertical __ H = (-.5 * g * 2.00^2) + 20.60 __ g is gravitational acceleration horizontal __ d = 19.0 * 2.00 b) H = 0 when the ball hits the ground 0 = (-.5 * g * t^2) + 20.60 solve for t to find the flight time
Physics
the k.e. is equal to the initial k.e. plus the potential energy from the drop k.e. = (.5 * .100 * 5.00^2) + (.100 * g * 10.0) g is gravitational acceleration
Math
a) 128 = (5 * 25) + (3 * 1) = (5 * 5^2) + (0 * 5^1) + (3 * 5^0) in base 5 __ 503 b)128 = 2^7 = (1 * 2^7) + (0 * 2^6) + (0 * 2^5) + (0 * 2^4) + (0 * 2^3) + (0 * 2^2) + (0 * 2^1) + (0 * 2^0) in base 2 __ 10000000 c) 128 = (10 * 12) + (8 * 1) = (10 * 12^1) + (8 * 12^0) in base 12...
Physics
the vertical component of the velocity (Vv)is __ 20.0 * sin(32.0º) the horizontal component (Vh) is __ 20.0 * cos(32.0º) the time (T) to max height is __ Vv / g a) Hmax = (-.5 g T^2) + (Vv * T) + 1.1 b) at Hmax, the ball has no vertical velocity; only horizontal (Vh)
PHYSICS
the time to accelerate to takeoff speed is __ 35 / 3.8 the average speed during acceleration is __ (0 + 35) / 2 the distance (length of runway) is speed times time
physics
0 = (-.5 * g * 5^2) + (10 * 5) + h g is gravitational acceleration
physics
Ha = -.5 g t^2 + 75 Hm = -.5 g t^2 + 30 t 75 = 30 t __ this is when substitute t into either equation to find where
Physics
a) [(2.25 m/s * 4.98 min) + (1.58 m/s * 2.18 min)] * 60 s/min b) (answer from a) / (4.98 min + 2.18 min) / (60 s/min)
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