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3 years ago

When the temperature of air rises, the amount of water needed for saturation

Physics

2 answers:

Oliga [24]3 years ago

7 0

It Increases. I just took a quiz with the same question.

Lemur [1.5K]3 years ago

6 0

<span>When the temperature of air rises, the amount of water needed for saturation increases. </span>

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A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?

Ipatiy [6.2K]

Answer:

Use equation for kinetic energy: Ek=mV²/2

m=700 kg

V=10m/s

Ek=700kg*100m²7s²/2

Ek=35000 J=35kJ

Explanation:

Hope this helps you

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5 0

3 years ago

Read 2 more answers

A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m$

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

$\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds$

The reaction time cannot be more than 1.69515 seconds

4 0

3 years ago

Which of the following is not affect by muscles a blood temperature b digestion c hair growth d speech

Murrr4er [49]

Answer:

Explanation:

i would think c would be correct because

blood temperature IS affected by muscles

digestion IS affected by muscles

speech IS affected by muscles

so therefore <em>hair growth</em> IS NOT affected by muscles

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3 0

3 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows: