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defon
2 years ago
7

Which of the following statements accurately describes the properties gases

Physics
2 answers:
Gwar [14]2 years ago
8 0

It may be, A change in the pressure of a gas results in a more significant change in volume than would occur in a liquid. I’m not for sure because I’m still studying this, but I think I correct but correct me if I’m wrong so that I can know and figure out what is the correct answer. Thank you

lana [24]2 years ago
5 0

2 they are easy to compress,

2 they expand to fill their containers,

3 they occupy far more space than the liquids or solids from which they form.

hope this helped!

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
ivolga24 [154]

Answer:

Half: 6 cm^2    Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

5 0
3 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a
wariber [46]
<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = \frac{V^{2} }{R}

=> R = \frac{V^{2} }{P}             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = \frac{V^{2} }{P}    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = \frac{12.0^{2} }{4}

R₁ = \frac{144}{4}

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = \frac{V^{2} }{P}    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = \frac{12.0^{2} }{4}

R₂ = \frac{144}{4}

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

\frac{1}{R_{X} } = \frac{1}{R_1} + \frac{1}{R_2}       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

\frac{1}{R_X} = \frac{1}{36} + \frac{1}{36}

\frac{1}{R_X} = \frac{2}{36}

Rₓ = \frac{36}{2}

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = \frac{11.7}{18}

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = \frac{0.3}{0.65}

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0
3 years ago
Read 2 more answers
Phương trình trạng thái tổng quát của khí lí tưởng diễn tả là​
tatyana61 [14]

Expla

pV/T=const goodluke

6 0
2 years ago
Find the speed vfinal of the joined cars after the collision. mastering physics
Tanya [424]
<span>Px = 0 Py = 2mV second, Px = mVcosφ Py = –mVsinφ add the components Rx = mVcosφ Ry = 2mV – mVsinφ Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²) and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) simplifying Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²) Vf = (V/3)âš((cosφ)² + (2 – sinφ)²) Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ)) Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ)) using the identity sin²(Ď)+cos²(Ď) = 1 Vf = (V/3)âš1 + 4 – 2sinφ) Vf = (V/3)âš(5 – 2sinφ)</span>
6 0
3 years ago
Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind
IrinaVladis [17]
Hello!

Most ocean waves obtain their energy and motion from the wind.

Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances. 

Have a nice day!


3 0
3 years ago
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