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3 years ago

1. The volume of a given mass of gas is 20cm when its

Physics

1 answer:

Pavel [41]3 years ago

5 0

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2 Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

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9. A balloon is filled with 1000 cm3 of a gas weighing 1000 g. Will it rise or fall when it is released?

torisob [31]

We can calculate the density of the balloon as follows:

$\rho=\frac{mass}{volume}=\frac{1000g}{1000cm^3}=\frac{1g}{cm^3}$

Therefore, the balloon will fall

Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.

3 0

1 year ago

As an object falls to the ground its E, is converted to

Andreyy89

Kinetic Energy I’m not 100% shure tho

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3 years ago

Read 2 more answers

How many significant digits are in the measurement 4,032,010 m/s?

choli [55]

Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.

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3 years ago

Read 2 more answers

A 3000-N force gives an object an acceleration of 15 m/s2. The mass of the object is

7nadin3 [17]

<h3>Answer:</h3>

200 kg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Newton's Law of Motions </u>

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion

Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)

Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u> </u>

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] F = 3000 N

[Given] a = 15 m/s²

[Solve] m = <em>x</em> kg

<u>Step 2: Solve for </u><em><u>m</u></em>

Substitute in variables [Newton's Second Law of Motion]: 3000 N = m(15 m/s²)
[Mass] [Division Property of Equality] Isolate <em>m</em> [Cancel out units]: 200 kg = m
[Mass] Rewrite: m = 200 kg

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2 years ago

An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which cont

erastova [34]

Answer:

$c_u=1540.5J/kg^{\circ}K$

Explanation:

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation $Q=mc\Delta T=mc(T_f-T_i)$ . The heat released by the unknown material is absorbed by water, so we have $Q_u=-Q_w$ , and we can write:

$m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})$

Since thermal equilibrium is reached we know that $T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K$ , where we have added $273^{\circ}$ to convert the temperature from Celsius to Kelvin, as <em>we must do</em>. Since we want the specific heat of the unknown material, we do:

$c_u=-\frac{m_wc_w(T_f-T_{wi})}{m_u(T_f-T_{ui})}$

Which for our values is:

$c_u=-\frac{(1.1kg)(4186J/kg^{\circ}K)((304^{\circ}K)-(294^{\circ}K))}{(0.49kg)((304^{\circ}K)-(365^{\circ}K))}=1540.5J/kg^{\circ}K$

3 0

3 years ago