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Alexeev081 [22]

3 years ago

13

Anybody wana play imvu or 2k

2 answers:

Yuliya22 [10]3 years ago

5 0

If you give me Brainly yeah

sergij07 [2.7K]3 years ago

4 0

Hmm I used to have those and then I deleted them. Sorry bud

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How to the eath turn

mrs_skeptik [129]

Explanation:

Earth rotates in prograde mation.As viewed from the north pole star Polaris.Earth turns counterclockwise,, the north pole is point in the northern,, Hemisphere where Earth's Axis of rotation meets it's surface

5 0

2 years ago

Read 2 more answers

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

Can anyone pls help me out in dis i am struggling in dis!

Step2247 [10]

The answer is C because all the other choices would cause disaster like pollute the water if you just throw it in the sink and it would smell terrible if you put it in the garbage etc...

3 0

2 years ago

Constant speed is the same thing as average speed

larisa86 [58]

No, they have different definitions

4 0

2 years ago

Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places

Olenka [21]

Answer:

The magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage $V_{out} = 17.33$

Resistance $R_{1} = 3$ kΩ

Voltage gain $A_{v} = 4.33$

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

$A_{v} = 1+\frac{R_{f} }{R_{1} }$

$4.33 = 1+ \frac{R_{f} }{3 k }$

$R_{f}$ ≅ 10 kΩ

For finding input voltage we use,

$A_{v} = \frac{V_{out} }{V_{2} }$

$V_{2} = \frac{17.33}{4.33}$

$V_{2} = 4$ V

The Phase of $V_{2}$ is zero because output voltage phase is 360°

Therefore, the magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

7 0

2 years ago