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inna [77]
3 years ago
7

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

3 x 10^-8 m^3/s; this water is replenished by the upward flow of sap through vessels in the trunk. This tree's trunk contains about 2000 vessels, each 100Mu m in diameter. What is the speed of the sap flowing in the vessels?
Physics
1 answer:
Gnoma [55]3 years ago
7 0

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV    .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

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Verify that the SI unit of impulse is the same as the SI unit of momentum.
lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

P = mv

Where:

P = momentum

m = mass      

v = velocity

The SI unit:

mass = kg\\ velocity = \dfrac{m}{s}

So the unit of momentum would be:

kg.\dfrac{m}{s}

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or kg.\dfrac{m}{s^{2} }

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = kg.\dfrac{m}{s^{2} } x s

or

\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}

You can then cancel out one s each from the numerator and denominator and you'll be left with:

kg.\dfrac{m}{s}

So then:

Momentum:                             Impulse

kg.\dfrac{m}{s}                                       kg.\dfrac{m}{s}

4 0
3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
3 years ago
Question 2 (Multiple Choice Worth 3 points)
zmey [24]
Surveys are considered the most reliable way to gather data
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The group of test subject that are NOT given the experimental treatment is called what?
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Answer:

The group that remains unaltered is called the control group.

5 0
3 years ago
PLEASE HELP!!!!
Aleks [24]
The answer is step by step 65
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2 years ago
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