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3 years ago

At which temperature could air hold the most water vapor?

Physics

1 answer:

Goshia [24]3 years ago

6 0

Explanation:

35 maybe hope it's right

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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

Tju [1.3M]

Answer:

v = 2.85 m/s

Explanation:

By the law of similarity of the triangle we can say

$\frac{height\: of\: man}{distance\: of \: man\: from \: spotlight} = \frac{height \:of \:image}{distance\: of \:wall\: from\: spotlight}$

here we know that

height of man = 2 m

let the distance of man from spotlight = x

distance of wall from spotlight = 12 m

height of image is let say "y"

so we will have

$\frac{2}{x} = \frac{y}{12}$

$y = \frac{24}{x}$

now we have

$\frac{dy}{dt} = \frac{24}{x^2}\frac{dx}{dt}$

here we know

$\frac{dx}{dt} = speed = 1.9 m/s$

x = 4 m

now we have

$\frac{dy}{dt} = \frac{24}{4^2}(1.9)$

$v = 2.85 m/s$

8 0

4 years ago

Match the half life and time information to the percentage of radioactive isotope left.

Kazeer [188]

Answer:

Explanation:

can i have brainliest pls im new

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3 years ago

Is this object showing acceleration for the first 2 seconds? explain your answer.

iVinArrow [24]

Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.

3 0

3 years ago

Friction, conduction and induction are three ways of creating a static charge on an object. Which method or methods will produce

blondinia [14]

The method that would produce the same charge is conduction only.

7 0

4 years ago

Read 2 more answers

A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically

Elanso [62]

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

Mass of softball = 0.220 kg

Speed = 5.5 m/s

(a). We need to calculate the velocity of the target ball

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0.220\times5.5+m_{2}\times0=0.220\times(-3.9)+m_{2}v_{2}$

$1.21=-0.858+m_{2}v_{2}$

$m_{2}v_{2}=2.068$ ....(I)

The velocity approach is equal to the separation of velocity

$u_{1}-u_{2}=v_{2}-v_{1}$

$5.5-0=v_{2}-(-3.9)$

$v_{2}=1.6\ m/s$

(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

$m_{2}\times1.6=2.068$

$m_{2}=\dfrac{2.068}{1.6}$

$m_{2}=1.29\ kg$

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

3 0

3 years ago