Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
Here it is given that speed of the dog will increase from 5 m/s to 11 m/s in order to cover the distance of 22 m
so here we can use kinematics to find the acceleration
![v_f = 11 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2011%20m%2Fs)
![v_i = 5 m/s](https://tex.z-dn.net/?f=v_i%20%3D%205%20m%2Fs)
![d = 22 m](https://tex.z-dn.net/?f=d%20%3D%2022%20m)
now we will have
![v_f^2 - v_i^2 = 2 a d](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202%20a%20d)
![11^2 - 5^2 = 2(a)(22)](https://tex.z-dn.net/?f=11%5E2%20-%205%5E2%20%3D%202%28a%29%2822%29)
![a = \frac{11^2 - 5^2}{44}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B11%5E2%20-%205%5E2%7D%7B44%7D)
![a = 2.18 m/s^2](https://tex.z-dn.net/?f=a%20%3D%202.18%20m%2Fs%5E2)
so here it will accelerate with a = 2.18 m/s^2
Answer:
false
Explanation:
The electrical force is much weaker than the force of gravity. False. Thomas Edison argued that our electrical system should use direct current because he felt that alternating current was dangerous.
Answer:
A.) False
B.) True
C.) True
D.) False
Explanation:
a. If the acceleration of an object remains constant, then its velocity is constant. False because if velocity is constant, acceleration will be equal to zero.
b. If the acceleration of an object moving along a line is always 0, then its velocity is constant. Yes. According to definition of acceleration, saying it is the rate of change of velocity. So if velocity is not changing, acceleration = 0
c. It is impossible for the instantaneous velocity at all times at b to equal the average velocity over the interval at b. Yes. Because of constant velocity
d. A moving object can have negative acceleration and increasing speed. False. Because when an object is coming to rest velocity is always reducing.
Answer:
flow of electric charges - current
opposition to the flow of an electric current - resistence
stored potential energy at the source of a circuit - voltag