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3 years ago

Calculate the final displacement from 0 if an object first moves 3 cm to the left and then 5 cm to the right.

2 answers:

5 0

So.. if it moves 3cm to the left and 5 cm to the right, you know that it takes 3cm to get it back to it's initial starting position, but it's been displaced.
So the final displacement is 2cm to the right.

antiseptic1488 [7]3 years ago

3 0

Answer: 2 cm to the right

Explanation:

The displacement is the shortest distance between initial and final position.

It can be calculated by subtracting the initial position from final position.

Let the sign be positive for direction towards right and negative for direction towards left.

The initial position is 0.

Firstly, object moves towards left = -3 cm

Then, the object moves towards right = 5 cm

Thus, the final position of the object is = -3 cm + 5 cm = +2 cm

Displacement = final position - initial position = + 2 cm - 0 = + 2 cm

Thus, the final displacement is 2 cm to right.

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sergij07 [2.7K]

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3 years ago

What conversion factor is used to convert 20 cm to m

Anastaziya [24]

The conversion factor you use is 100 cm = 1 m.
You can divide 20 by 100 to get the answer.
20 cm/100 cm =.2 m
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7 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t

vampirchik [111]

The new period is D) √2 T

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

$\texttt{ }$

$\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}$

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

$T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}$

$T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}$

$T : T_2 = \sqrt{L} : \sqrt{2L}$

$T : T_2 = 1 : \sqrt{2}$

$\boxed {T_2 = \sqrt{2}\ T}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

3 0

3 years ago

Read 2 more answers

You are in charge of designing a movie stunt for a new Fast and Furious film. In this scene a stunt person has to jump from a br

blondinia [14]

Answer:

103.5 meters

Explanation:

Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.

The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:

h = Ut + 1/2gt^2

since the person will fall under gravity, g = 9.8 m/s^2

Also, let assume that the person jump from rest, then, U = 0

Substitute h, U and g into the formula above

22.5 = 1/2 * 9.8 * t^2

22.5 = 4.9t^2

t^2 = 22.5/4.9

t^2 = 4.59

t = $\sqrt{4.592}$

t = 2.143 seconds

From definition of speed,

speed = distance /time

Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.

48.3 = distance / 2.143

distance = 48.3 * 2.143

distance = 103.5 m

Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?

3 0

3 years ago