Answer:
The answer to your question is: d = 0 m, it does not move
Explanation:
Data
vo = 20 m/s
a = -1 m/s2
t = 40 s
d = ?
Formula
d = vot + (1/2)at²
Substitution
d = (20)(40) + (1/2)(-1)(40)²
d = 800 - 800
d = 0 m It suggest that it does not move.
I hope it can help you
Answer:

Explanation:
What problem says can be written mathematically as:

Where:

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

The derivative of the function space as a function of time is equal to a function of speed as a function of time.
The velocity vector is given by the vector sum of the velocities of both axes.

If you notice any mistake in my english, please let me know, because I am not native.
Answer:
Aluminium, Nickel, Cobalt and Iron plus varying levels of Copper, Titanium and Niobium.
MAGNETIC: Alni, Alcomax, Hycomax, Columax, and Ticonal.
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