Why might the 1990 earthquake and volcanic eruption in Luzon be related?

A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0

3 years ago

reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0

Verizon [17]

Answer:

Displacement: 2.230 km Average velocity: 1.274 $\frac{km}{h}$

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

0.916 km due east is an horizontal direction and cane be seen as direction towards the negative side of X-axis.

0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
3.52 km in a direction of 49.7° has components on X and Y axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑ $Sx = Ax -0.916$ ⇒ ∑ $Sx = [tex]3.52cos(49.7) - 0.916$