N a scientific investigation, what is the name for a prediction that can be tested?

Explain a lever and a pivot in a full sentence

worty [1.4K]

Answer:

The lever is a movable bar that pivots on a fulcrum attached to a fixed point. The lever operates by applying forces at different distances from the fulcrum, or a pivot. As the lever rotates around the fulcrum, points farther from this pivot move faster than points closer to the pivot.

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4 0

2 years ago

A package is dropped from a helicopter moving upward at 15 m/s

daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s² × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996 + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

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7 0

1 year ago

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.866 s. How much time

Iteru [2.4K]

<h2>Entire trip takes 1.22 seconds.</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 0.866 s

Substituting

s = ut + 0.5 at²

s = 0 x 0.866 + 0.5 x 9.81 x 0.866²

s = 3.68 m

Halfway is 3.68 m

Total height = 2 x 3.68 = 7.36 m

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = ?

Displacement, s = 7.36 m

Substituting

s = ut + 0.5 at²

7.36 = 0 x t + 0.5 x 9.81 x t²

t = 1.22 s

Entire trip takes 1.22 seconds.

7 0

2 years ago

A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the

ehidna [41]

Answer: 170.67 N

Explanation:

Given

Mass of skier is $m=82.9\ kg$

Height of the inclination is $h=20\ m$

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

$\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N$

Thus, the horizontal friction force is 170.67 N.

7 0

3 years ago

A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in

Kryger [21]

The average power is $3.0\cdot 10^6 W$

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$