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3 years ago

Which of the following is an example of a primary source?

Physics

1 answer:

Oduvanchick [21]3 years ago

8 0

Answer:

i think its no c.

a textbook written by a scientist. because its a first hand data.

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When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?

Reika [66]

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

8 0

2 years ago

Read 2 more answers

A crane lifts a 425 kg steel beam vertically upward a distance of 614.4 m

Maru [420]

PHYSICS

*not sure about the answer but here we go*

Mass = 425 kg

distance = 614.4 m

acceleration = 1.8 m/s²

Answer :

Count Force first.

$f \: = m \: \times a$

F = 425 × 1,8

F = 765 N ✅

Now let's count Work.

$w \: = \: f \: \times s$

W = 765 × 614.4

W = 470016 J ✅

7 0

2 years ago

The speed of the wooden bar is changed so that the bar hits the water fewer times each second.

Hitman42 [59]

Answer:

c - the frequency of the waves decreases

3 0

2 years ago

if monochromatic light passes from water into air with an angle of incidence of 35 degress which characteristic of light will re

Kamila [148]

Frequency will be same

7 0

3 years ago

On a hot summer day, the density of air at atmospheric pressure at 31.5°C is 1.2074 kg/m3.

mario62 [17]

To solve this problem, it is necessary to apply the ideal Gas equations, as well as the calculation equations of the weight difference, which under the comparison of two values.

By definition we know that the ideal gas equation is given by the equation,

PV = nRT

Where,

P = Pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = number of moles

Our values are given by

$P = 1.1013*10^5Pa$

$T = 31.5\°C = 304.5K$

$\rho = 1.2074kg/m^3$

PART A) Using this previous equation we can find the number of moles per Volume, that is

$PV = nRT$

$\frac{n}{V} = \frac{P}{RT}$

Replacing with our values

$\frac{n}{V} = \frac{1.013*10^5}{(8.314)(304.5)}$

$\frac{n}{V} = 40.014mol/m^3$

PART B ) We can calculate the number of moles of 1m^3 through Avogadro number, then

$M = nA$

$M = 40.014*(2.88*10^{-2})$

$M = 1.1524Kg$

Therefore in $1m^3$ there are 1.1524Kg of Gas.

PART C ) Density can be defined as the proportion of mass in a specific quantity of Volume, then

$\rho_2 = \frac{M}{V}$

$\rho_2 = \frac{1.1524}{1}$

$\rho_2 = 1.1524kg/m^3$

The difference of percentage then is

$\Delta \rho = \frac{\rho_1-\rho_2}{\rho_2}*100\%$

$\Delta \rho = \frac{1.2074-1.1524}{1.1524}*100\%$

$\Delta \rho = 0.0477*100\%$

$\Delta \rho = 4.77\%$

YES, because as the percentage is less than 10%, the calculated value agrees with the stated value.

5 0

2 years ago