Answer:
0.0000000010 joules
Explanation:
Amount : 1540.3 nanojoules (nJ)
Equals : 0.0 joules (J)
Answer:
2 m/s²
Explanation:
the equations of motion are
S= ut +½at²
v² = u²+ 2as
v = u + at
s = (u+v)/2 × t
From the parameters given
u = 0m/s this is because it starts from rest
Distance (s) = 9m
Time (t) = 3s
Based on this the first equation would be used
s = ut + ½at²
Input values
9 = 0×3 + ½ × a x 3²
9 = 0 + 9a/2
9 = 4.5a
Divide both sides by 4.5
a = 9 / 4.5 m/s²
a = 2 m/s²
I hope this was helpful, please mark as brainliest
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.
![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20x%28%5Ctext%7Bvertical%7D%29%20%5C%5C%5B0.5em%5D%20%26%3D%20-%5Cfrac%7B1%7D%7B2%7D%5C%2C%20g%20%5Ccdot%20t%5E%7B2%7D%20%2B%20v_0%5Ccdot%20t%5C%5C%5B0.5em%5D%20%26%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-2%7D%20%5Ctimes%20%287.5%5C%3B%20%5Crm%20s%29%5E%7B2%7D%20%5C%5C%20%26%20%5Cquad%20%5Cquad%20%2B%200%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-1%7D%20%5Ctimes%207.5%5C%3B%20s%20%5C%5C%5B0.5em%5D%20%26%3D%20-281.25%5C%3B%20%5Crm%20m%5Cend%7Baligned%7D)
.
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 
.
