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3 years ago

Suppose an object is in free fall. Each second the object falls

Physics

1 answer:

Dmitry [639]3 years ago

6 0

Answer:

The correct answer option is b) a larger distance than in the second before.

Explanation:

Supposing that an object is in free fall, each second the object would fall a larger distance than in the second before.

In Physics, free fall is the state of motion of an object where the only force acting upon it is the force of gravity.

So for a object experiencing free fall, with every passing second it covers a larger distance in comparison to the previous second.

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A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m

Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0 (1)

2.070k + Φ = 70.53. (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0

4 years ago

light waves are first transmitted through the ________ at the front of the eye and enter an opening called the ________ before s

viva [34]

The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.

Light waves can be regarded as moving energy which contains microscopic particles known as photons.

The vision of the eye can be completed through the light wave passing through the components of the eyes and this process goes thus;

Light will move through the (cornea) which is situated at the front area of the eyes into lens.

Then both the cornea and the lens give room for the focusing of the light rays to the retina which is situated at the back of the eye .

Then through the help of the cells in the retina, the light will be absorbed and then be converted to electrochemical impulses and then transfer it to the brain as well as optic nerve.

Therefore, light wave are form of tiny microscopic particles.

brainly.com/question/19734585?referrer=searchResults

8 0

3 years ago

A rabbit with a mass of 2.0 kg accelerates at 2.0 meters per second squared. Find the net force of the rabbit

scZoUnD [109]

Answer:

Explanation:

Force = mass x acceleration

Given

Mass = 2.0kg

Acceleration = 2.0m/s^2

Force = 2.0 x 2.0

= 4N

5 0

4 years ago

Read 2 more answers

Un contenedor de 1800 N está en reposo sobre un plano inclinado a 28°, el coeficiente de fricción entre el contenedor y el plano

babunello [35]

Answer:

F = 1480.77N

Explanation:

In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:

$F-Wsin\alpha-F_r=0\\\\F-Wsin\alpha-N\mu cos\alpha=0\\\\F-Wsin\alpha-W\mu cos\alpha=0$ (1)

F: required force = ?

W: weight of the container = 1800N

N: normal force = weigth

α: angle of the incline = 28°

g: gravitational acceleration = 9.8m/s^2

μ: coefficient of friction = 0.4

You solve the equation (1) for F and replace the values of the other parameters:

$F=W(sin\alpha+\mu cos\alpha)\\\\F=(1800N)(sin28\°+(0.4)cos28\°)=1480.77N$

The required force to push the container for the incline with a constant velocity is 1480.77N

8 0

3 years ago

100 POINTS 100 POINTS 100 POINTS!!!!! HELP PLEASE I DON'T KNOW WHAT TO DO!!!!!

Westkost [7]

Answer:

block 2 or 4

because of the distribution of weight and force being applied to the object

3 0

3 years ago

Read 2 more answers