A. Genus and B. Species are also major levels of classification.
Answer:
27%
Explanation:
15.999 divided by 58.32 = .27433128
Move the decimal place over 2 places.
27%
Answer:
Approximately
. (Assuming that the drag on this ball is negligible, and that
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.) - Vertical: constant downward acceleration at
, starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given:
. Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of
, accelerated downwards at
, and reached the ground after
.
Apply the SUVAT equation
to find the vertical displacement of this ball.
.
In other words, the ball is
below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be
.
Answer:
4.02 s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 35°
Initial velocity (u) = 50 m/s
Acceleration due to gravity (g) = 10 m/s²
Time of flight (T) =?
The time of flight of the arrow can be obtained as follow:
T = 2uSineθ / g
T = 2 × 35 × Sine 35 / 10
T = 70 × 0.5736 / 10
T = 7 × 0.5736
T = 4.02 s
Therefore, the time taken for the arrow to return is 4.02 s
Answer:
40 m/s
Explanation:
If you multiply the acceleration with time you get the average speed.