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3 years ago

During which lunar phase is the moon's far side entirely lit

Physics

2 answers:

Alla [95]3 years ago

7 0

The phase where the moons far side is entirely lit is called the full moon

yulyashka [42]3 years ago

4 0

<span>During which lunar phase is the moon's far side entirely lit?</span><span>

"Near side entirely lit" means we see the whole side facing us. The moon looks like a complete circle.

Hope This Helped! :3</span>

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Chemical name and a common name for NaOH

garri49 [273]

Chemical Name: Sodium Hydroxide

7 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>

<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>

$W_A = -P_2(V_2 - V_1)$

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

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

$\texttt{ }$

<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

4 years ago

A car accelerates for 10 seconds. During this time, the angular

bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, $\omega_i$ = 10 rad/s

final angular velocity, $\omega_f$ = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

$a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2$

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0

3 years ago

What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

3 years ago

If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the

Yuri [45]

Answer:

<u>The energy change would be 46kJ</u>

<u>The energy would be absorbed</u>

Explanation:

The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.

Hence:

<em>Energy change</em> = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.

4 0

3 years ago