Answer:
b
Explanation:
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.
This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.
Answer: coefficient of static friction
= 0.31
Explanation: Since they negotiate the curve without skidding, the frictional force (F1) equals the centripetal force (F2).
F1= uN
F2 = M*(v²/r)
M is the combined mass 450kg
V is the velocity 18m/s
r is the radius 106m
N is the normal reaction 4410N
u is the coefficient of static friction
Making u subject of the formula we have that,
u = {450*(18²/106)} /4410
=1375.47/4410
=0.31
NOTE: coefficient of friction is dimensionless. It as no Unit.
Answer:
t= 137.5 s
Explanation:
So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another
<u>Slopes:</u>
Runner A: y = 7.50x + 55
Runner B: y = 7.90 x
sooooo
7.50 x + 55 = 7.90 x
- 7.50 x - 7.50 x
55 = .40 x
55/.40 = .40 x / .40
x = 137.5 s
t= 137.5 s
7.50 * 137.5 + 55 =1086.25 m
7.90 * 137.5 = 1086.25 m
Speed = distance / time
If you input your numbers into this equation you will be able to find the cyclists average speed
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 
.
