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ozzi
3 years ago
12

Data that shows consistent, regular, repetitive form displays a pattern. True False

Physics
2 answers:
anyanavicka [17]3 years ago
6 0
I think the answer is True
Agata [3.3K]3 years ago
3 0
I think the statement is true.
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A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Vinvika [58]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

4 0
4 years ago
Amy gets up from her seat on a train to move closer to the front. Just as she begins to walk forward, the train stops at a stati
Ivahew [28]
Have you ever been in a bus, a train, or a car ?
What happens to you when it stops suddenly ?

A body in motion continues in motion unless
an external force acts on it and makes it stop.

Amy's body keeps moving forward when the train stops.
She pitches forward, and if she doesn't reach out and grab
a seat or a seated person, she may lose her footing and fall
on her face.

Choice - 'A' is a very good explanation.
The other choices aren't.

'C' is a good hunch, but it only applies to her feet.
The rest of her keeps going.

'D' is nonsense.  There are no mysterious forces of
'repulsion' or 'attraction' on the train.
5 0
3 years ago
Read 2 more answers
(50 POINTS) HELP!!!!!! WILL GIVE BRAINLIEST
zepelin [54]
C) During chemical reaction, total mass remains the same.

This is because due to the law of conservation of mass, stating that in a chemical reaction, neither mass will be created nor destroyed. So, the total mass will not change when chemical reactions occur.
7 0
3 years ago
Calculate the change in internal energy (ΔE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.0
Alexxandr [17]

Explanation:

The given data is as follows.

         q = -25.0 kJ,     Pressure (P) = 1.50 atm

   \Delta V = (12 - 6) L = 6 L

Therefore, product of pressure and change in volume will be as follows.

             P \Delta V = 1.50 atm \times 6 L

                                = 7.5 L atm

                                = 7.5 \times 101.3

                                = 759.75 J

Now, we will calculate the change in internal energy as follows.

                   \Delta E = q + w

                                = q + P \Delta V

                                = -25000 kJ + 759.75 J

                                 = 24240.25 J

or,                              = 24.240 kJ      (as 1 kJ = 1000 J)

Thus, we can conclude that the change in internal energy (\Delta E) for a system is 24.240 kJ.

7 0
4 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
4 years ago
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