Answer:
Speed of the speeder will be 28 m/sec
Explanation:
In first case police car is traveling with a speed of 90 km/hr
We can change 90 km/hr in m/sec
So 
Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m
After that car is accelerating with
for 7 sec
So distance traveled by car in these 7 sec

So total distance traveled by police car = 224 m
This distance is also same for speeder
Now let speeder is moving with constant velocity v
so 
v = 28 m/sec
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg
Now,new weight with g = 6m/s^2
=m×g' (here g' is new acceleration of the new planet)
= 10×6=60N
The horizontal velocity<span> of a projectile is </span>constant<span> (a never </span>changing<span> in value), There is a </span>vertical<span>acceleration caused by gravity; its value is 9.8 m/s/s, down, The </span>vertical velocity<span> of a projectile </span>changes<span> by 9.8 m/s each second, The </span>horizontal<span> motion of a projectile is independent of its </span>vertical<span> motion.</span>
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)