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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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The force between two interacting charges is 9.0 × 10-5 newtons. They are kept 1 meter apart. If the magnitude of one charge is
Oliga [24]

The magnitude of other charge will be 1 × 10⁻² coulomb

The formula of electrostatic force is

Electrostatic force = K q1 q1 / r²

where k is the coulomb's constant whose value is 9 × 10⁹

q1 and a2 are the magnitude of charges

and r is the distance between them

magnitude of the force given to us is 9.0 × 10⁻⁵ newtons

magnitude of one charge = 1.0 × 10⁻⁶ coulomb

Force = K q1 q2 / r²

9.0 × 10⁻⁵ = ( ( 9 × 10⁹ ) × ( 1.0 × 10⁻⁶ ) × q2 ) / 1

9.0 × 10⁻⁵ =  9 × 10³ × q2  

10⁻² = q2  

Charge on q2 is 1 × 10⁻² coulomb

So the magnitude of the second charge is came out to be 1 × 10⁻² coulomb after applying the formula of electrostatic force.

Learn more about electrostatic force here:

brainly.com/question/17692887

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8 0
2 years ago
A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters a
vitfil [10]

Answer:

<u>At 268.82°C</u> volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

V ∝ T

\frac{V}{T}=k

where k is a constant.

For two samples of gases, the law can be given as:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

From the data given:

V_1=5.5\ l

T_1=25\ \°C =(273+25)K= 298 K

V_2=10\ l

We need to find T_2.

Plugging in values in the formula.

\frac{5.5}{298}=\frac{10}{T_2}

Multiplying both sides by T_2.

T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2

\frac{5.5}{298}T_2={10}

Multiplying both sides by \frac{298}{5.5}

\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}

T_2=541.82\ K

T_2=541.82\ K-273\ K = 268.82\°C

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

7 0
3 years ago
Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat o
timama [110]

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ

So, 425 kJ of thermal energy is severed.

6 0
3 years ago
Block 2 of mass 1.00 kg is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 20
son4ous [18]

Answer:

x = 0.327 m

Explanation:

Block 2 of mass 1.00 kg is at rest

spring constant = 200 N/m            

Block 1 of mass 2 kg moving at 4 m/s

m₁ v₁ = (m₁ + m₂)V                                

2 x 4 = (2 + 1) V                                                      

V = 2.67 m/s                                                        

loss of kinetic energy = gain elastic potential energy

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2                  

\dfrac{1}{2}\times (3)\times 2.67^2 = \dfrac{1}{2}\times 200 \times x^2

x = 0.327 m

hence, the spring compressed distance is equal to x = 0.327 m

8 0
3 years ago
Ello<br><br>NY1 ZINDA HERE xD<br><br>WHAT IS MOLE ??​
Julli [10]

Answer:

yes I am

how are you

have a great day

8 0
3 years ago
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