A group of physics students hypothesize that for an experiment they are performing, the speed of an object sliding down an incli

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

Position or composition

<span>a. To determine the velocity at which the camera strikes the ground: v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera impacts the ground with a speed of 42 m/s. b. To calculate the duration it takes for the camera to reach the bottom: y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds
       
The camera descends for 11.4 seconds before hitting the ground.</span>

Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = = = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as

where u is the velocity at the lowest point. Consequently, the modified equation is

=

=

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v =

=

=

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.