12. Which of these is more dense?

How do you calculate elastic potential energy

Nimfa-mama [501]

U=1/2kx2

This image sums it up

5 0

2 years ago

Determine qual é a carga elétrica de um corpo que possui 1 milhão de partículas?

MaRussiya [10]

Answer:

This can be translated to:

"find the electrical charge of a body that has 1 million of particles".

First, it will depend on the charge of the particles.

If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C

Then the total charge of the body will be:

Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C

If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)

Q = 1.6*10^-13 C

But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.

7 0

2 years ago

Which of the following is not the one of the influencing factors when it comes to body type

serg [7]

Where are the answer choice ?

3 0

2 years ago

Read 2 more answers

12 points , just need help with this question

tigry1 [53]

Answer:f 30

Explanation: I am not really sure but try this

3 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