Answer:
The magnitude of the force is 1.29*10^-3N in the positive x direction
Explanation:
In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:
(1)
where:
You replace the expression for U into the equation (1) and solve for F:
![F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3](https://tex.z-dn.net/?f=F%28x%29%3D-%5Cfrac%7Bd%7D%7Bdx%7D%5BAx%5E4%5D%3D-4Ax%5E3)
(2)
The force on the particle, for x = -0.080m is:
The magnitude of the force is 1.29*10^-3N in the positive x direction
<h3>One reason we should know what the directions of the forces acting on an object is so we can know if we have to add or subtract. same = add together, opposite = subtract from each other. Also if we don't pay attention to the direction the Net force will be wont be accurate. There will be factors that will upset the calculation.So we must know the direction of the two forces because we have to know if we are adding or subtracting and if the answer is accurate. </h3>
<em>I hope this helps!.</em>
Answer:
POSITIVE
Explanation:
The acceleration of an object is given by the rate of change of velocity.
u and v are initial and final velocities
If the final velocity is more than that of the initial velocity, the acceleration of the object is positive. It means positive acceleration occurs when an object speeds up. Hence, the correct option is (a) "positive".
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
