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nika2105 [10]
3 years ago
11

Which of the following is NOT a physical mechanism that causes air to rise?A) orographic liftingB) frontal wedging C) adiabatic

coolingD) convergence
Physics
1 answer:
liq [111]3 years ago
5 0

Answer:

Option C

Explanation:

Adiabatic cooling systems function similarly to dry cooling systems, but with the incorporation of pre-cooling pads; running water over pre-cooling pads and drawing air through the pads depresses the ambient dry bulb of the incoming air. The depressed dry bulb allows for greater system heat rejection.

he result of this is that adiabatic systems are highly effective in hot, dry environments, while using less water than traditional evaporative units. Adiabatic units also deliver the required cooling capacity in a smaller footprint and/or lower fan motor horsepower than a completely dry cooler/condenser.

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Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert this
Paladinen [302]
1 m = 1 000 000 ym

converted other way we can say that:

1 ym = 10^{-6} m

Now, since we have ym^2 which is ym*ym which means:
1 ym^2 = (10^{-6}) ^2  =  10^{-12} m

we have 1,5 ym^2 which means that answer is:
1.5* 10^{-12} m
8 0
3 years ago
Question:<br> the distance of the car can move by velocity 15 m/s for time 5s is :<br> Physics
tangare [24]

Answer:

c) 2.05

Explanation:

7 0
2 years ago
A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the
Yuri [45]

Ans:  Let d is the distance from height to our eyes.

<span>Applying the Pythagoras theorem, we  get,</span>

<span>Check image: </span>

7 0
3 years ago
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