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2 years ago

How is speed motion related

Physics

1 answer:

GrogVix [38]2 years ago

8 0

Speed is a measure of how fast something is in motion. Hope this helps!

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Which of the following is NOT an indicator of a physical change?

S_A_V [24]

Answer:

1. C. The change is easily reversible

2. A. a physical change

Explanation:

Happy Holidays

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2 years ago

Why does Energy appear continuous in everyday life?

lara31 [8.8K]

Answer:

The first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another. For example, turning on a light would seem to produce energy; however, it is electrical energy that is converted. Nothing happens to the energy. It does not change form, since energy has no form. ... If the energy was moving, it gets stored or re-transmitted elsewhere. Using energy means controlling its movement, rather than consuming it.

8 0

3 years ago

What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90

Anestetic [448]

Answer:

i think its the third one or second one

4 0

2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

A vessel contains 1 mol of O2 and 2mol of He.what is the value of'Cp/Cv ' of the mixture?

BlackZzzverrR [31]

Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1

Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2

Now using formula of effective degree of freedom of mixture, (f), we have:

f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)

We know f=11/6,
substituting the value in above relation, we have:

Cv= R. 11/3*2
= R. 11/6

Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R

Therefore, Cp/Cv = 17/11

6 0

2 years ago