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Nady [450]
3 years ago
9

The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume

of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.
Express your answer using one significant figure.

P = ? Pa
Physics
1 answer:
timurjin [86]3 years ago
6 0

Answer:

P = 1 x 10⁸ Pa

Explanation:

given,

radius = 2.0 ×10⁻¹⁰ m

Temperature

T = 300 K

Volume of gas molecule =

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3}\pi (2\times 10^{-10})^3

 V = 33.51 x 10⁻³⁰ m³

we know,

P  V = 1 . k T

k = 1.38  x 10⁻²³ J/K

P(33.51 x 10⁻³⁰) = 1 . (1.38  x 10⁻²³) x 300

P = 1.235 x 10⁸ Pa

for 1 significant figure

P = 1 x 10⁸ Pa

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velikii [3]
M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).

The momentum before colision is equal with the momentum after colision:

m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem

=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
7 0
3 years ago
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
JulijaS [17]

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

5 0
3 years ago
5. Which unit of electricity does the work in a circuit?
Pachacha [2.7K]

Answer:

Volt

Explanation:

Voltage is what makes electric charges move. ... Voltage is also called, in certain circumstances, electromotive force (EMF). Voltage is an electrical potential difference, the difference in electric potential between two places. The unit for electrical potential difference, or voltage, is the volt.

The ohm is defined as an electrical resistance between two points of a conductor when a constant potential difference of one volt, applied to these points, produces in the conductor a current of one ampere, the conductor not being the seat of any electromotive force.

The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). ... In terms of SI base units, the coulomb is the equivalent of one ampere-second. Conversely, an electric current of A represents 1 C of unit electric charge carriers flowing past a specific point in 1 s.

An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.

8 0
3 years ago
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

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2 years ago
You toss a bowling ball straight up into the air with a speed of 2.1 m/s. How long does it take the bowling ball to reach its hi
lara [203]

Time taken by the bowling ball to reach its highest point= 0.214 s

initial velocity= Vi=2.1 m/s

Final velocity= Vf=0 as the velocity at the highest point is zero.

acceleration= g= -9.8 m/s²

using the kinematic equation Vf= Vi + at

0= 2.1 + (-9.8)t

t= -2.1/-9.8

t=0.214 s

Thus the time taken by the bowling ball to reach its highest point is 0.214 s

6 0
3 years ago
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