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4 years ago

How to Identify Newtons Three Laws in my Own Words?

Physics

1 answer:

ipn [44]4 years ago

5 0

I understand the struggle. Here's a way to do it.

Identify Newton's Three Laws, and let it be important that you understand what you're reading. If there's a passage, read the passage/text and look for any important detail you should add to your piece, and add any information you have learned from the passage/text and put it into your own words. You should make sure that the information you've written down is consistent with the passage/text and that your phrasing is not too close to the original.

-I anticipate that this would help you.

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Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each

zloy xaker [14]

Answer:

* thermal energy

potential energy,

potential energy

Explanation:

The ship has various types of energy,

* thermal energy. This energy is associated with the temperature of the coal, the hotter the greater its internal energy,

* potential energy this energy is stored in the constituent atoms within carbon

* potential energy. It is due to the configuration of the system, in this case the sun heats the coal

5 0

3 years ago

An object oscillates with an angular frequency of 8.0 rad/s. At t = 0, the object is at x0 = 4 cm with an initial velocity v0 =

KatRina [158]

Answer:

$\phi = 0.66 rad$

$A = 5.06 cm$

Explanation:

We have here a simple harmonic motion, so the equation of the position in this motion is:

$x(t)=Acos(\omega t+\phi)$ (1)

A: Amplitude

ω: angular frequency

φ: phase constant

If we take the derivative of x with respect to t from (1), we can find the velocity equation of this motion:

$v(t)=\frac{dx(t)}{dt}=-A\omega sin(\omega t+\phi)$ (2)

Let's evaluate (1) and (2) in t=0.

$x(0)=Acos(\phi)$ (3)

$v(0)=-A\omega sin(\phi)$ (4)

Dividing 4 by 3 we have:

$\frac{v(0)}{x(0)}=-\omega tan(\phi)$

$\phi = tan^{-1}(\frac{-v(0)}{\omega x(0)})$

$\phi = 0.66 rad$

Now, using (3) we can find the amplitude.

$A = \frac{x(0)}{cos(\phi)} = 5.06 cm$

I hope it helps!

6 0

4 years ago

580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum(m=3)?

snow_tiger [21]

Answer:

56.3

Explanation:

5 0

4 years ago

Read 2 more answers

The work function of an element is the energy required to remove an electron from the surface of the solid. The work function fo

vichka [17]

Answer:

λ = 548.7 nm

Explanation:

Hi!

First we want to know how much energy we need to remove 1 electron from the surface of the solid:

218.1 kJ/mol => 218 100 J / (6.022 x 10^23) electrons

= 3.621 x 10^-19 J/electron

That is we need 3.621 x 10^-19 J to remove one electron

Now we can calculate the wavelength that a photon must have in order to have this energy:

E = (hc) / λ

λ = (hc) / Ε

where

h = 6,626070150(69) ×10 -34 Js (wikipedia)

c = 3 x10^8 m/s

hc = 1.987 x 10^-25 Jm

Therefore:

λ = ( 1.987 x 10^-25 /3.621 x 10^-19 ) m = 5.487 x 10^-7 m

λ = 548.7 nm

4 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

4 years ago