Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,

Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

Time the parachutist falls without friction is 3.19 seconds

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds


Now the initial velocity of the last half height will be the final velocity of the first half height.

Since the height are equal


Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

Magnitude of acceleration is -43200 ft/s²
315g/95gmol-1
3.315 moles of MgCl2
Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 

k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be


An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.