In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.
Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.
In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Therefore the correct option is therefore A
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This column actually appears due to the body parts reflecting the light of the sun or moon to the observer.
If the body is not perfectly smooth, only parts of this body would be at the correct position to reflect the light of the sun or moon to the observer.
If the body was perfectly smooth, it would act like a mirror where you can see the full image of the sun or the moon reflected on it.
A, C, and D are true when a light spot appears, so those are false.
The answer is therefore B.
the resulting wave speed=62.7 m/s
Explanation:
The speed of a wave is given by
v=
T= tension, v= velocity= 58 m/s
when the tension is increased by 17%, new tension= 1.17 T
V'=1.082 V
V'=1.082 (58)
V'=62.7 m/s
What is missing is the density of salt water. Let's denote it ρ.
<span>The pressure at depth d then is: </span>
<span>p = p0 + ρ g d </span>
<span>, where p0 is the atmospheric pressure at the surface of the sea. We assume it is equal to the 1 atm inside the sub. </span>
<span>Hence the pressure difference between inside and outside of the window is: </span>
<span>Δp = ρ g d </span>
<span>The force on the window is this pressure difference, multiplied by the area of the circular window of diameter D. The latter equals </span>
<span>A = ¼ π D². </span>
<span>So </span>
<span>F = p A = ¼ π D ²ρ g d. </span>
<span>This should be less than Fmax. </span>
<span>Hence d < Fmax / (¼ π D ²ρ g ). </span>
