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3 years ago

Convert 5.6kg to grams

Physics

2 answers:

OLEGan [10]3 years ago

4 0

Answer:

my pleasure

Explanation:

5.6kg = 5600 grams

multiply the mass value by 1000

Kruka [31]3 years ago

4 0

Answer: 5.6kg = 5600 grams

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If the current through a bulb in a series circuit is 0.4 A and the potential difference across the bulb is 3 V then what is the

sergey [27]

Resistance of the bulb is 7.5 ohms

Explanation:

In a circuit, according to Ohm's Law, the potential difference (voltage) across an element is directly proportional to the current flowing through it.

That is, V ∝ I

⇒ V = IR, where R is the constant of proportionality called the resistance.

Here, current, I = 0.4 A and potential difference, V = 3 V

Resistance, R = V/I = 3/0.4 = 7.5 ohms

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3 years ago

A gas in a cylinder is held at a constant pressure of 1.80 * 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3. The in

Lilit [14]

Answer:

The work done is $W=-0.9*10^5J$ and it is negative because the exterior is doing work on the gas.

The value of the heat flow is $Q=2.3*10^5J$ and is directed out of the gas, as this is being cooled and compressed.

Explanation:

The work done by the gas can be written as

$W=\int\limits^ {V_{c}}_{V_{h}}{P}\, dV$

and all the data is given in the problem, so

$W=P(V_{c}-V_{h})=-0.9*10^5J$

is the work necessary to compress the gas in the cylinder, wich is negative because the exterior is doing work on the gas.

Now, from first law of thermodynamics, we have that

$\Delta U=Q+W$

<em>where $\Delta U$ is the internal energy difference, $W$ is the calculated work, and $Q$ is the heat flow that we wish to know</em>, then

$Q=\Delta U-W=1.4*10^5J+0.9*10^5J=2.3*10^5J$

And the direction of the heat flow is outside of the gas, as it is being cooled and compressed.

Finally, it doesn't matter whether the gas is ideal or not, because we never use the Ideal Gas hypotesis, that $PV=nRT$ , and $P$ is constant.

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3 years ago

A solution is a homogeneous mixture of one or more solutes dissolved in a solvent. A specific volume of solvent is only able to

Luba_88 [7]

<h3><u>Answer;</u></h3>

C. Supersaturated

<h3><u>Explanation</u>;</h3>

Solutions are homogeneous mixtures that are created by mixing a solute and a solvent. Solute is the substance present in smaller amounts that dissolves in a solvent such as water which is the substance present in larger amount.
A solution, can be<u> unsaturated, saturated or supersaturated. An unsaturated solution</u> is a solution that contains less solute that can be dissolved, it doesn't contain the maximum amount of solute.
<u>A saturated solution</u> is a solution containing the maximum amount of solute that can be dissolved at a given temperature. Any additional solute will remain undissolved in the container.
<u>A supersaturated solution</u> is a solution created when a solution is carefully cooled because it contains more solute than the solubility allows.

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3 years ago

Read 2 more answers

In a certain region of space, the electric field is constant in direction (say horizontal, in the x direction), but its magnitud

horsena [70]

Answer:

8.3 x 10⁻⁷ C

Explanation:

Electric flux will enter the face at x=0 and exit at face x= 25 m

On the other faces , field lines are parallel so no flux will enter or exit .

Flux entering the face at x = 0

= electric field x face area

= 560 x 25 x 25 = 350000 weber

Flux exiting the face at x = 25

= 410 x 25 x25

= 256250 weber

Net flux exiting from cube ( closed face )

350000 - 256250 = 93750 web

Apply gauss'es theorem

Q / ε = Flux coming out

Q is charge inside the closed cube

Q / ε = 93750

Q = 8.85 x 10⁻¹² x 93750

= 8.3 x 10⁻⁷ C

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3 years ago

Which of the following properties did Rutherford use in his experiment?

Ronch [10]

<u>Answer</u>:

(C) The positive charge of the alpha particles and the negative charge of the electrons are the properties Rutherford used in his experiment

<u>Explanation</u>:

In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.

He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.

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4 years ago