The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

$f=\mu\,* N = \mu * m g\, cos(\theta)$

and $w_{//}= m\,g\,sin(\theta)$

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

$f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)$

For this, the sole additional tool needed is a protractor for angle measurement.

7 0

2 months ago

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

inna [3103]

The height measures 69.68 m

Explanation:

Given data

Before striking the ground = 46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

Thus, the height is 69.68 m

6 0

2 months ago

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of

ValentinkaMS [3465]

<span>3.834 m/s. To solve this problem, we must ensure that the centripetal force equals or exceeds the gravitational force acting on the object. The formula for centripetal force is F = mv^2/r while the equation for gravitational force is F = ma. Since the mass (m) cancels out in both equations, we can equate them, leading to a = v^2/r. Now, inserting the given values (where the radius is half the diameter) allows us to find v: 9.8 m/s^2 <= v^2/1.5 m, which simplifies to 14.7 m^2/s^2 <= v^2. Therefore, we find that the minimum velocity required is 3.834057903 m/s <= v. Thus, the necessary speed is 3.834 m/s.</span>

4 0

2 months ago