Answer:
The driver hits the stationery dog because the applied force is less than required force
Explanation:
Kinetic energy will be given by
where m is the mass of the vehicle and v is the speed/velocity of the vehicle.
Substituting 800 Kg for m and 20 m/s for v we obtain
Frictional force by vehicle pads is given by
where d is the distance moved
Substituting 160000 for KE and 50 m for d we obtain
Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N
Answer:
D
Explanation: It makes the most sense. Plz mark brainliest
Answer: a) 127 eV; b) there is no change of kinetic energy.
Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field. This means the electric force do work in this trayectory and then the protons increased changes its speed.
If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then they can not move to higher potential if any external force does work on the system.
In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons can not move to lower potential region (V=-40V).
Answer : The value of the constant for a second order reaction is, 
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ?
t = time = 17s
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
= final concentration = 0.0981 M
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:
Therefore, the value of the constant for a second order reaction is,
Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up
