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denis23 [38]
3 years ago
11

Assume that you are in a space station conducting fluid mechanics experiments in zero gravity. You perfectly coat a sphere and a

long cylinder with a thin layer of the same fluid and thickness. If the sphere and cylinder have the same radius, will the fluid layers on the two objects have the same pressure
Physics
1 answer:
erastova [34]3 years ago
5 0

Answer:

Explanation:

From the Question operation is

-zero gravity

-same fluid thickness

-same radius

Generally the force is given by f=ma=\frac{p}{a}

=> g=m/s^2=0

=> As acceleration varies directly to pressure

=>p=0

Generally it is shown that the fluid layer on the two objects will have same pressure.

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The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.

In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.

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3 0
2 years ago
A marble on a frictionless track, starting from point A in the drawing, is projected down the curved runway. (This means that th
EleoNora [17]

Answer:

v = 4.4 m / s

Explanation:

Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.

Starting point. Point A

          Em₀ = K + U = ½ m v2 + mg y_a

point B.

          Em_f = U = m g y

the energy is conserved

         Em₀ = Em_f

         ½ m v² + mg y_a = m g y

        ½ m v² = m g (y -y_a)

         v = \sqrt {2g ( y - y_a)}

         In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m

       v = \sqrt{ 2 \  9.8 ( 5 -4)}

       v = 4.4 m / s

4 0
3 years ago
An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that
zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e=2\times 1.6\times 10^{-19} C

1 e=1.6\times 10^{-19} C

Mass of alpha particle=m=6.64\times 10^{-27} kg

Potential difference,V=1.97\times 10^6 V

Magnetic field,B=3.49 T

a.Speed of alpha particle=v=\sqrt{\frac{2 qV}{m}}

By using the formula

v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}

v=1.38\times 10^7 m/s

b.Magnetic force,F=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N

F=1.5\times 10^{-11} N

c.Radius of circular path, r=\frac{mv^2}{F}

r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}

r=0.084 m

5 0
3 years ago
Daily point..........d​
Alina [70]

Answer:

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Explanation:

4 0
3 years ago
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Chemical weathering in a sentence
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