Answer: 1 m/s
Explanation:
We have an object whose position 
is given by a vector, where the components X and Y are identified by the unit vectors 
and 
(where each unit vector is defined to have a magnitude of exactly one):
![r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=r%3D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
On the other hand, velocity is defined as the variation of the position in time:
This means we have to derive :
![\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5Cfrac%7Bd%7D%7Bdt%7D%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
This is the velocity vector
And when 
the velocity vector is:
This is the velocity vector at 2 seconds
However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed 
:
Finally:
This is the speed of the object at 2 seconds
Velocity = (displacement) / (time)
Displacement = straight-line distance between start-point and end-point
If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
F4.0
Explanation:
To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.
