Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00

At x = 1.00 m

= 4J
Kinetic energy = (1/2)mv²

= 18J
Total energy will be =
4J + 18J = 22J
At x = 5

= -0.24J
Kinetic energy =

= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2

= 3.33 m/s
b) magnitude of force when x = 5.0m



At x = 5.0 m


= 0.016N
Answer:
434 Hz
Explanation:
According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.
In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.
The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.
Answer:
If you are simply looking for the X component then the most applicable formula from the choices given is Tx + Ux+ Vx. This means that you will add all x-components. For example: If a man walking along the x-axis walks 10 meters to the right, 5 back and 2 meters forward, what is the resultant vector?
Imma go with A.
Hope this helps:)