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3 years ago

Help me please, it's so hard

Physics

2 answers:

fomenos3 years ago

8 0

Answer:

4.58×10²³ atoms

5.94×10⁻²¹ J

1340 m/s

Explanation:

Use ideal gas law to find moles of gas.

PV = nRT

(1.266 atm × 101300 Pa/atm) (4/3 π (0.15 m)³) = n (8.31451 J/mol/K) (14 + 273) K

n = 0.760 mol

Use Avogadro's number to find number of atoms.

(0.760 mol) (6.02214×10²³ atom/mol) = 4.58×10²³ atoms

Average kinetic energy per molecule is:

KE = 3/2 kT

KE = 3/2 (1.38066×10⁻²³ J/K) (14 + 273) K

KE = 5.94×10⁻²¹ J

RMS speed of each atom is:

KE = 1/2 mv²

5.94×10⁻²¹ J/atom = 1/2 (0.004 kg/mol) (1 mol / 6.02214×10²³ atom) v²

v = 1340 m/s

valentina_108 [34]3 years ago

6 0

1340 m/s is the answer and I got it how he got it^ but I’m too lazy to right it all down lol

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In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number p

Brums [2.3K]

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

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2 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

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Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

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3 years ago

A 1.2 nf parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nf when the gap is filled

Damm [24]

Answer:

2.5

Explanation:

The capacitance of a parallel-plate capacitor filled with dielectric is given by

$C=kC_0$

where

k is the dielectric constant

$C_0$ is the capacitance of the capacitor without dielectric

In this problem,

$C_0=1.2 nF$ is the capacitance of the capacitor in air

$C=3.0 nF$ is the capacitance with the dielectric inserted

Solving the equation for k, we find

$k=\frac{C}{C_0}=\frac{3.0 nF}{1.2 nF}=2.5$

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3 years ago

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You have 2 objects flying towards you. One of them is havier with lower velocity and the other one is lighter but has higher vel

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Answer: I would say the object with the Lower velocity because Lighter with Higher velocity makes it heavy, velocity is how heavy something is so the lighter it is the less difficult it will be to catch.

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3 years ago

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