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topjm [15]
3 years ago
13

Explain why atmospheric pressure changes as atmospheric depth changes

Physics
1 answer:
Pavel [41]3 years ago
7 0

The force of the collisions creates pressure on the container. ... Explain why atmospheric pressure changes as depth changes. Atmospheric pressure increases as depth increases because at lower levels of the atmosphere, there is more air above that is being pulled down by gravitational force.

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What is the Acceleration of a car traveling 25 m/s and in 5 seconds<br> traveling 65 m/s? *
WARRIOR [948]

Answer:

8m/s²

Explanation:

a = change in v/t

a = (65-25)/5

a = 8

7 0
3 years ago
Cual de las escalas de temperatura es la mas antigua
IrinaVladis [17]

Answer:

the translation I got for this question is

Which of the temperature scales is the oldest?

Explanation:

and i searched for it and got this=

Fahrenheit scale

6 0
2 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
Why is detergent mixed in a toothpaste?
babymother [125]

because detergent has to clean dirt outa clothes and tooth pste also can help get stains outa cloths and when you put those two mixtures tofgether it can make the ultimate cleaning mixture

8 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
3 years ago
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