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Rina8888 [55]
3 years ago
14

The capacity of the air to hold water vapor: Group of answer choices 1. decreases with an increase in temperature.2. increases w

ith a decrease in temperature.3. increases with an increase in temperature.4. increases with an increase in pressure.
Physics
2 answers:
abruzzese [7]3 years ago
8 0

Answer:

Option (3)

Explanation:

The capacity of air to hold moisture or water content is directly proportional to the temperature of the air. This means that, with the increasing temperature, the air is capable of holding more amount of water. This results in a decreasing amount of relative humidity. When the temperature of the air increases and the water vapor content increases simultaneously, it slowly starts to condense and thereby leads to the occurrence of precipitation.

Thus, the correct answer is option (3).

Nezavi [6.7K]3 years ago
5 0

Answer:

3. increases with an increase in temperature.

Explanation:

The air more water vapor at higher temperatures because at higher temperatures the air expands and the inter-molecular space increases so the room for water molecules increases.

Warm air keeps the water molecules warm and prevents them from condensing.

The air can hold the moisture only upto its saturation quantity after which the precipitation occurs in the form of rain, snow, hail, sleet etc.

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Given data

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*The final velocity of the first glider is v_2 = -1 m/s

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Applying the law of conservation of momentum as

\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}

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\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}

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Solving a series circuit, did I do this correctly? ​
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  • The total resistance in the circuit is 16 Ohms.
  • The total current in the circuit is 0.5 Ampere.
  • The current at R_1 is 0.5 Ampere.
  • The current at R_3 is 0.5 Ampere.
  • The voltage drop atR_1 is 4 volts.
  • The voltage drop at R_2 is 2.5  volts.
  • The voltage drop at R_3 is 1.5 volts.
  • The total power consumed by the circuit is 4watts
  • The power consumed at R_1 is 2 watts
  • The power consumed at R_2 is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

  • The total resistance in the circuit  = R_{eq}

R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega

  • The total current in the circuit = I

V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A (Ohm's law)

  • For series combinations, the current in each resistor remains the same.

So, the current in R_1, R_2 \&R_3:

I_1= I_2= I_3=I=0.5 A\\

  • The voltage drop across at R_1 = V_1

The current across  R_1 = I = 0.5 A

V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V

  • The voltage drop across at R_2 =V_2

The current across  R_2 = I = 0.5 A

V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V

  • The voltage drop across at R_3 = V_3

The current across  R_3 = I = 0.5 A

V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V

  • The total power consumed by circuit:

P= V\times I \\\\= 0.5 A\times 8 V = 4 watt

  • Power consumed at R_1:

P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt

  • Power consumed at R_2:

P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt

  • Power consumed at R_3:

P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt

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