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3 years ago

The diagram below shows some waves traveling along a rope.

Physics

1 answer:

Harrizon [31]3 years ago

3 0

-- Well first of all, the waves are <em>transverse waves</em>. The fibers in the rope, and the hand that generates the waves, are moving up and down, but the wave is moving to the right, towards the wall. These directions are perpendicular.

-- Later on, after the waves reflect from the wall and travel back toward the hand, there are going to be <em>standing waves</em> on the rope. But this is probably beyond the scope of the question.

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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

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The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

Komok [63]

Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

brainly.com/question/12198573

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mr Goodwill [35]

Answer:

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