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Alexeev081 [22]
3 years ago
6

How are evaporation and transpiration similar?

Physics
2 answers:
julsineya [31]3 years ago
8 0

Answer:

A

Explanation:

Iteru [2.4K]3 years ago
7 0
The answer is A. They are both processes in which water is changed into water vapor.
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The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
2 years ago
What is a planet’s period of rotation?
Annette [7]


The answer is the letter "C" ( I have honors science I am good at this type of stuff )

Hope I helped :) ( ask me for help when u need it :)

5 0
3 years ago
Read 2 more answers
What is an independent variable
Iteru [2.4K]

Answer:

An independent variable is a variable that is manipulated to determine the value of a dependent variable. The dependent variable is what is being measured in an experiment or evaluated in a mathematical equation and the independent variables are the inputs to that measurement.

Explanation:

8 0
3 years ago
Read 2 more answers
A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is in
Gnoma [55]

Answer:

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

        P = F / A

Where the area of ​​a circle is

       A = π r²

E radius is half the diameter

      r = d / 2

      A = π d² / 4

We replace

    P = F 4 / π d²2

    P₁ = 397 4 /π  0.058²

    P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

   P₁ = P₂ + ½ ρ v₂²

  v₂ = √ (P1-P2) 2 / ρ

  v₂ = √ [(1.50-1.013) 10⁵ 2/1000]

  v₂ = 97.4 m / s

6 0
3 years ago
A small plane flies 37.0 km in a direction 45° north of east and then flies 28.0 km in a direction 25° north of east.
Karo-lina-s [1.5K]

Answer:

d= 64.1 km θ = 36.4º

Explanation:

a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.

If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)

In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).

In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after  completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:

x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km

Appying the same considerations for the y-coordinate, we have:

y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km

Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:

d = \sqrt{x1^{2}+y1^{2}} =\sqrt{(51.6km)^{2}+(38km)^{2}} =64.1 km

The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:

tg θ =\frac{y1}{x1} = \frac{38km}{56.1km} =0.736

⇒ θ = 36.4º North of East (counterclockwise from the east axis).

7 0
2 years ago
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