<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,

From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values

t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 

h = 124.694 m
So height of ramp = 124.694 m
Answer:
a single closed path of electrical components including a voltage source
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Answer:
<h3>The answer is 500 kg</h3>
Explanation:
The mass of the object can be found by using the formula

v is the velocity
KE is the kinetic energy
From the question we have

We have the final answer as
<h3>500 kg</h3>
Hope this helps you